MCQ 1011 Mark
$\int_{}^{} {{e^{2x}}( - \sin x + 2\cos x)\;dx = } $
- A
${e^{2x}}\sin x + c$
- B
$ - {e^{2x}}\sin x + c$
- C
$ - {e^{2x}}\cos x + c$
- ✓
${e^{2x}}\cos x + c$
AnswerCorrect option: D. ${e^{2x}}\cos x + c$
d
(d)$\int_{}^{} {{e^{2x}}( - \sin x + 2\cos x)\,dx} $
$ = - \int_{}^{} {{e^{2x}}\sin x\,dx} + 2\int_{}^{} {{e^{2x}}\cos x\,dx} $
$ = {e^{2x}}\cos x - 2\int_{}^{} {{e^{2x}}\cos x\,dx + 2\int_{}^{} {{e^{2x}}\cos x\,dx + c} } $
$ = {e^{2x}}\cos x + c.$
Aliter : $\int_{}^{} {{e^{2x}}(2\cos x - \sin x)\,dx} = {e^{2x}}\cos x + c$
$\left\{ \because \,\,\,\int_{{}}^{{}}{{{e}^{kx}}\left\{ k\,f(x)+{f}'(x) \right\}dx={{e}^{kx}}f(x)+c} \right\}$
View full question & answer→MCQ 1021 Mark
$\int_{}^{} {[f(x)\,g''(x) - f''(x)\,g(x)]\,dx} $=
- A
$\frac{{f(x)}}{{g'(x)}}$
- B
$f'(x)g(x) - f(x)g'(x)$
- ✓
$f(x)g'(x) - f'(x)g(x)$
- D
$f(x)g'(x) + f'(x)g(x)$
AnswerCorrect option: C. $f(x)g'(x) - f'(x)g(x)$
c
(c)$\int_{}^{} {[f(x)\,g''(x) - f''(x)\,g(x)]\,dx} $
$ = \int_{}^{} {f(x)\,g''(x)\,dx} - \int_{}^{} {f''(x)\,g(x)\,dx} $
$ = \left( {f(x)\,g'(x) - \int_{}^{} {f'(x)g'(x)\,dx} } \right) - \left( {g(x)\,f'(x) - \int_{}^{} {g'(x)\,f'(x)\,dx} } \right)$
$ = f(x)\,g'(x) - f'(x)\,g(x).$
View full question & answer→MCQ 1031 Mark
The value of $\int {\frac{{\log x}}{{{{(x + 1)}^2}}}dx} $ is
- ✓
$\frac{{ - \log x}}{{x + 1}} + \log x - \log \,(x + 1)$
- B
$\frac{{\log x}}{{\left( {x + 1} \right)}} + \log x - \log \,(x + 1)$
- C
$\frac{{\log x}}{{x + 1}} - \log x - \log \,(x + 1)$
- D
$\frac{{ - \log x}}{{x + 1}} - \log x - \log \,(x + 1)$
AnswerCorrect option: A. $\frac{{ - \log x}}{{x + 1}} + \log x - \log \,(x + 1)$
a
(a)$\int {\frac{{\log x}}{{{{(x + 1)}^2}}}dx = \int {\log x\,{{(x + 1)}^{ - 2}}} } dx$
$ = \log x.\left\{ { - {{(x + 1)}^{ - 1}}} \right\}$$ - \int {\frac{1}{x}.\{ - {{(x + 1)}^{ - 1}}\} dx} $
$ = \frac{{ - \log x}}{{(x + 1)}} + \int {\frac{1}{{x(x + 1)}}dx} $$ = \frac{{ - \log x}}{{(x + 1)}} + \int {\left[ {\frac{1}{x} - \frac{1}{{x + 1}}} \right]dx} $
$ = \frac{{ - \log x}}{{x + 1}} + \log x - \log (x + 1)$.
View full question & answer→MCQ 1041 Mark
$\int_{}^{} {32{x^3}{{(\log x)}^2}dx} $ is equal to
- ✓
${x^4}\{ 8{(\log x)^2} - 4(\log x) + 1\} + c$
- B
${x^3}\{ {(\log x)^2} + 2\log x\} + c$
- C
${x^4}\{ 8{(\log x)^2} - 4\log x\} + c$
- D
$8{x^4}{(\log x)^2} + c$
AnswerCorrect option: A. ${x^4}\{ 8{(\log x)^2} - 4(\log x) + 1\} + c$
a
(a) Let $I = \int_{}^{} {32{x^3}{{(\log x)}^2}} dx = 32\int_{}^{} {{x^3}{{(\log x)}^2}dx} $
$ = 32\,\left[ {{{(\log x)}^2}\int_{}^{} {{x^3}dx - \int_{}^{} {\left( {\frac{d}{{dx}}{{(\log x)}^2}\int_{}^{} {{x^3}dx} } \right)\,dx} } } \right]$
$ = 32\,\left[ {{{(\log x)}^2}.\frac{{{x^4}}}{4} - \int_{}^{} {2\log x.\frac{1}{x}.\frac{{{x^4}}}{4}dx} } \right]$
$ = 32\left[ {{{(\log x)}^2}\frac{{{x^4}}}{4} - \frac{1}{2}\int_{}^{} {{x^3}\log x\,dx} } \right]$
$ = 32\left[ {\frac{{{{(\log x)}^2}{x^4}}}{4} - \frac{1}{2}\left( {\frac{{\log x.{x^4}}}{4} - \int_{}^{} {\frac{1}{x}.\frac{{{x^4}}}{4}} {\rm{ }}dx} \right)} \right]$
$ = 32\left[ {\frac{{{{(\log x)}^2}{x^4}}}{4} - \frac{1}{2}\left( {\frac{{{x^4}\log x}}{4} - \frac{1}{4}.\frac{{{x^4}}}{4}} \right)} \right] + c$
$ = 8\,\left[ {{{(\log x)}^2}{x^4} - \frac{1}{2}\left( {{x^4}\log x - \frac{{{x^4}}}{4}} \right)} \right] + c$
$ = 8{x^4}\left[ {{{(\log x)}^2} - \frac{{\log x}}{2} + \frac{1}{8}} \right] + c$
$ = {x^4}[8{(\log x)^2} - 4\log x + 1] + c$.
View full question & answer→MCQ 1051 Mark
$\int_{}^{} {{{\sin }^{ - 1}}(3x - 4{x^3})dx = } $
- A
$x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + c$
- B
$x{\sin ^{ - 1}}x - \sqrt {1 - {x^2}} + c$
- C
$2[x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} ] + c$
- ✓
$3[x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} ] + c$
AnswerCorrect option: D. $3[x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} ] + c$
d
(d) Put $x = \sin \theta \Rightarrow dx = \cos \theta \,d\theta ,$ therefore
$\int_{}^{} {{{\sin }^{ - 1}}(3x - 4{x^3})} \,dx = \int_{}^{} {{{\sin }^{ - 1}}(\sin 3\theta )\cos \theta \,d\theta } $
$ = \int_{}^{} {3\theta \cos \theta \,d\theta } = 3\left\{ {\theta \sin \theta - \int_{}^{} {\sin \theta \,d\theta } } \right\}$
$ = 3\left\{ {\theta \sin \theta + \cos \theta } \right\} + c = 3\left\{ {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} } \right\} + c.$
View full question & answer→MCQ 1061 Mark
$\int_{}^{} {{{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}dx = } $
- A
$x{\tan ^{ - 1}}x + c$
- B
$x{\tan ^{ - 1}}x - \log (1 + {x^2}) + c$
- C
$2x{\tan ^{ - 1}}x + \log (1 + {x^2}) + c$
- ✓
$2x{\tan ^{ - 1}}x - \log (1 + {x^2}) + c$
AnswerCorrect option: D. $2x{\tan ^{ - 1}}x - \log (1 + {x^2}) + c$
d
(d) Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta ,$ then
$\int_{}^{} {{{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\,dx} = \int_{}^{} {{{\tan }^{ - 1}}\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \,\,\,{\sec ^2}\theta \,d\theta $
$ = \int_{}^{} {{{\tan }^{ - 1}}(\tan 2\theta ){{\sec }^2}\theta \,d\theta } = \int_{}^{} {2\theta {{\sec }^2}\theta \,d\theta } $
$ = 2\left[ {\theta \tan \theta - \int_{}^{} {\tan \theta \,d\theta } } \right]$$ = 2x{\tan ^{ - 1}}x - \log ({x^2} + 1) + c.$
View full question & answer→MCQ 1071 Mark
$\int_{}^{} {\frac{{{{\sin }^{ - 1}}x}}{{{{(1 - {x^2})}^{3/2}}}}\;dx = } $
- ✓
$\frac{x}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x + \frac{1}{2}\log (1 - {x^2}) + c$
- B
$\frac{x}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x - \frac{1}{2}\log (1 - {x^2}) + c$
- C
$\frac{1}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x - \frac{1}{2}\log (1 - {x^2}) + c$
- D
$\frac{1}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x + \frac{1}{2}\log (1 - {x^2}) + c$
AnswerCorrect option: A. $\frac{x}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x + \frac{1}{2}\log (1 - {x^2}) + c$
a
(a) Put $t = {\sin ^{ - 1}}x \Rightarrow \sin t = x \Rightarrow \cos t\,dt = dx,$ then
$\int_{}^{} {\frac{{{{\sin }^{ - 1}}x}}{{{{(1 - {x^2})}^{32}}}}\,dx} = \int_{}^{} {t{{\sec }^2}t\,dt = t\tan t + \log \cos t + c} $
$ = {\sin ^{ - 1}}x\tan ({\sin ^{ - 1}}x) + \log \cos ({\sin ^{ - 1}}x) + c$
$ = \frac{x}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x + \frac{1}{2}\log (1 - {x^2}) + c.$
View full question & answer→MCQ 1081 Mark
$\int_{}^{} {\frac{{x{{\tan }^{ - 1}}x}}{{{{(1 + {x^2})}^{3/2}}}}\;dx = } $
- A
$\frac{{x + {{\tan }^{ - 1}}x}}{{\sqrt {1 + {x^2}} }} + c$
- ✓
$\frac{{x - {{\tan }^{ - 1}}x}}{{\sqrt {1 + {x^2}} }} + c$
- C
$\frac{{{{\tan }^{ - 1}}x - x}}{{\sqrt {1 + {x^2}} }} + c$
- D
AnswerCorrect option: B. $\frac{{x - {{\tan }^{ - 1}}x}}{{\sqrt {1 + {x^2}} }} + c$
b
(b) Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta ,$ then
$\int_{}^{} {\frac{{x{{\tan }^{ - 1}}x}}{{{{(1 + {x^2})}^{32}}}}\,dx} = \int_{}^{} {\frac{{\theta \tan \theta {{\sec }^2}\theta \,d\theta }}{{{{(1 + {{\tan }^2}\theta )}^{32}}}}} $
$ = \int_{}^{} {\theta \sin \theta \,d\theta } = - \theta \cos \theta + \sin \theta + c$
$ = \frac{x}{{\sqrt {{x^2} + 1} }} - {\tan ^{ - 1}}x\frac{1}{{\sqrt {{x^2} + 1} }} = \frac{{x - {{\tan }^{ - 1}}x}}{{\sqrt {1 + {x^2}} }} + c$.
View full question & answer→MCQ 1091 Mark
$\int_{}^{} {{x^5}.{e^{{x^2}}}dx = } $
- ✓
$\frac{1}{2}{x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c$
- B
$\frac{1}{2}{x^4}{e^{{x^2}}} + {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c$
- C
$\frac{1}{2}{x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} - {e^{{x^2}}} + c$
- D
AnswerCorrect option: A. $\frac{1}{2}{x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c$
a
(a) Put ${x^2} = t \Rightarrow 2x\,dx = dt,$ then
$\int_{}^{} {{x^5}{e^{{x^2}}}dx} = \frac{1}{2}\int_{}^{} {{t^2}{e^t}dt} = \frac{1}{2}\left[ {{e^t}{t^2} - 2\int_{}^{} {t{e^t}dt} } \right] + c$
$ = \frac{{{t^2}{e^t}}}{2} - \left[ {t{e^t} - {e^t}} \right] + c = \frac{1}{2}{x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c.$
View full question & answer→MCQ 1101 Mark
If an antiderivative of $f(x)$ is ${e^x}$ and that of $g(x)$ is $\cos x,$then $\int {f(x)\cos x\,dx} + \int {g(x){e^x}dx = } $
- A
$f(x)g(x)+c$
- B
$f(x)+g(x)+c$
- ✓
${e^x}\cos x + c$
- D
$f(x) / g(x) + c$
AnswerCorrect option: C. ${e^x}\cos x + c$
c
(c)$\int {f(x)\cos xdx + \int {g(x){e^x}dx} } $
$ = \int {{e^x}\cos xdx + \int {( - \sin x){e^x}dx} } $
$ = \frac{{{e^x}}}{2}(\cos x + \sin x) - \frac{{{e^x}}}{2}(\sin x - \cos x) + c$
$ = \frac{{{e^x}}}{2}(2\cos x) + c$$ = {e^x}\cos x + c$.
View full question & answer→MCQ 1111 Mark
$\int_{}^{} {\sin \sqrt x } \;dx = $
- A
$2[\sin \sqrt x - \cos \sqrt x ] + c$
- ✓
$2[\sin \sqrt x - \sqrt x \cos \sqrt x ] + c$
- C
$2[\sin \sqrt x + \cos \sqrt x ] + c$
- D
$2[\sin \sqrt x + \sqrt x \cos \sqrt x ] + c$
AnswerCorrect option: B. $2[\sin \sqrt x - \sqrt x \cos \sqrt x ] + c$
b
(b) Put $\sqrt x = t \Rightarrow \frac{1}{{2\sqrt x }}\,dx = dt \Rightarrow dx = 2t\,dt,$ then
$\int_{}^{} {\sin \sqrt x \,dx} = 2\int_{}^{} {t\sin t\,dt} = 2( - t\cos t + \sin t) + c$
$ = 2(\sin \sqrt x - \sqrt x \cos \sqrt x ) + c.$
View full question & answer→MCQ 1121 Mark
The value of $\int {{{\sec }^3}x\,\,dx} $ will be
- ✓
$\frac{1}{2}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$
- B
$\frac{1}{3}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$
- C
$\frac{1}{4}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$
- D
$\frac{1}{8}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$
AnswerCorrect option: A. $\frac{1}{2}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$
a
(a) Let $I = \int {{{\sec }^3}xdx} $$ = \int {\sec x{{\sec }^2}xdx} $
$ \Rightarrow I = \sec x\tan x - \int {\sec x\,{{\tan }^2}x\,dx} $
$ \Rightarrow I = \sec x\tan x - \int {\sec x\,({{\sec }^2}x - 1)dx} $
$ \Rightarrow I = \sec x\tan x - \int {{{\sec }^3}x\,dx} + \int {\sec x\,dx} $
$ \Rightarrow \;I = \sec x\tan x - I + \log \,(\sec x\, + \tan x\,)$
$ \Rightarrow 2I = \sec x\tan x + \log (\sec x + \tan x)$
$ \Rightarrow I = \frac{1}{2}[\sec x\tan x + \log (\sec x + \tan x)]$.
View full question & answer→MCQ 1131 Mark
If $I = \int_{}^{} {{e^x}\sin 2x\;dx} $, then for what value of $ K$ , $KI = {e^x}(\sin 2x - 2\cos 2x) + $constant
Answerc
(c) $I = \int_{}^{} {{e^x}\sin 2x\,dx} = \sin 2x\,.\,{e^x} - 2\int_{}^{} {\cos 2x\,.\,{e^x}dx} $
$ = \sin 2x\,.\,{e^x} - 2\cos 2x\,.\,{e^x} - 4\int_{}^{} {{e^x}\sin 2x\,dx} $
$ \Rightarrow 5I = {e^x}(\sin 2x - 2\cos 2x) + $constant
Equating the given value, we get $K = 5.$
View full question & answer→MCQ 1141 Mark
$\int_{}^{} {x\sqrt {2x + 3} } \;dx = $
- ✓
$\frac{x}{3}{(2x + 3)^{3/2}} - \frac{1}{{15}}{(2x + 3)^{5/2}} + c$
- B
$\frac{x}{3}{(2x + 3)^{3/2}} + \frac{1}{{15}}{(2x + 3)^{5/2}} + c$
- C
$\frac{x}{2}{(2x + 3)^{3/2}} + \frac{1}{6}{(2x + 3)^{5/2}} + c$
- D
AnswerCorrect option: A. $\frac{x}{3}{(2x + 3)^{3/2}} - \frac{1}{{15}}{(2x + 3)^{5/2}} + c$
a
(a)$\int_{}^{} {x{{(2x + 3)}^{1/2}}dx} $
$ = x\frac{{{{(2x + 3)}^{3/2}}}}{{3/2}}\frac{1}{2} - \int_{}^{} {\frac{{{{(2x + 3)}^{3/2}}}}{{3/2}}\frac{1}{2}\,dx + c} $ $ = \frac{1}{3}x{(2x + 3)^{3/2}} - \frac{1}{3}\int_{}^{} {{{(2x + 3)}^{3/2}}dx + c} $
$ = \frac{1}{3}x{(2x + 3)^{3/2}} - \frac{1}{{15}}{(2x + 3)^{5/2}} + c.$
View full question & answer→MCQ 1151 Mark
$\int_{}^{} {u\frac{{{d^2}v}}{{d{x^2}}}dx - \int_{}^{} {v\frac{{{d^2}u}}{{d{x^2}}}dx = } } $
AnswerCorrect option: A. $u\frac{{dv}}{{dx}} - v\frac{{du}}{{dx}} + c$
a
(a) $\int_{}^{} {u\frac{{{d^2}v}}{{d{x^2}}}\,dx} - \int_{}^{} {v\frac{{{d^2}u}}{{d{x^2}}}\,dx} $
$ = u\frac{{dv}}{{dx}} - \int_{}^{} {\frac{{du}}{{dx}}\,.\,\frac{{dv}}{{dx}}\,dx - v\frac{{du}}{{dx}} + \int_{}^{} {\frac{{dv}}{{dx}}\,.\,\frac{{du}}{{dx}}\,dx + c} } $
$ = u\frac{{dv}}{{dx}} - v\frac{{du}}{{dx}} + c.$
View full question & answer→MCQ 1161 Mark
If $f(x) = g(x)$, then the value of $\int_{}^{} {f'(x)} \;.\;g(x)\;dx$ is
- A
$\frac{1}{2}{\left\{ {g(x)} \right\}^2} + c$
- B
${\left\{ {g(x)} \right\}^2} + c$
- C
$\frac{1}{2}{\left\{ {f(x)} \right\}^2} + c$
- ✓
$A$ or $C$ both
AnswerCorrect option: D. $A$ or $C$ both
d
(d) Given that $f(x) = g(x)$. Now $\int_{}^{} {f'(x)\,g(x)\,dx} $
$ = g(x)\int_{}^{} {f'(x)\,dx} - \int_{}^{} {\left[ {g'(x)\int_{}^{} {f'(x)} \,dx} \right]} \,dx + c$
$ = g(x)\,f(x) - \int_{}^{} {g'(x)\,f(x)\,dx} + c$
$ \Rightarrow \int_{}^{} {f'(x)\,g(x)\,dx} = g(x)\,f(x) - \int_{}^{} {f'(x)\,g(x)\,dx + c} $
Since $f(x) = g(x) \Rightarrow f'(x) = g'(x)$ …..$(i)$
From $(i),$ $2\int_{}^{} {f'(x)\,g(x)\,dx} = g(x)\,f(x) + c$
Hence $\int_{}^{} {f'(x)\,g(x)\,dx} = \frac{1}{2}{\left\{ {g(x)} \right\}^2} + c$
Also $\int_{}^{} {f'(x)\,.\,g(x)\,dx} = \int_{}^{} {f'(x)\,.\,f(x)\,dx} $ = $\frac{1}{2}{\{ f(x)\} ^2} + c$.
Aliter : $\int_{}^{} {f'(x)\,.\,g(x)\,dx} = \int_{}^{} {f'(x)\,.\,f(x)} \,dx$
$ = \int_{}^{} {t\,dt} $, $\left\{ {f(x) = t \Rightarrow f'(x)\,dx = dt} \right\}$
$ = \frac{{{t^2}}}{2} + c = \frac{{{{\left\{ {f(x)} \right\}}^2}}}{2} + c = \frac{{{{\left\{ {g(x)} \right\}}^2}}}{2} + c$.
View full question & answer→MCQ 1171 Mark
$\int_{}^{} {\frac{1}{{{x^2}}}\log ({x^2} + {a^2})dx = } $
- A
$\frac{1}{x}\log ({x^2} + {a^2}) + \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c$
- ✓
$ - \frac{1}{x}\log ({x^2} + {a^2}) + \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c$
- C
$ - \frac{1}{x}\log ({x^2} + {a^2}) - \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c$
- D
AnswerCorrect option: B. $ - \frac{1}{x}\log ({x^2} + {a^2}) + \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c$
b
(b) $\int_{}^{} {\frac{1}{{{x^2}}}\log ({x^2} + {a^2})\,dx} = \int_{}^{} {{x^{ - 2}}\log ({x^2} + {a^2})\,dx} $
$ = \frac{{ - \log ({x^2} + {a^2})}}{x} + \int_{}^{} {\frac{{2x}}{{({x^2} + {a^2})}}.\frac{1}{x} + c} $
$ = \frac{{ - \log ({x^2} + {a^2})}}{x} + \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c.$
View full question & answer→MCQ 1181 Mark
$\int_{}^{} {\log x(\log x + 2)\;dx = } $
AnswerCorrect option: A. $x{(\log x)^2} + c$
a
(a) $I = \int_{}^{} {\log x(\log x + 2)\,dx} $
Put $\log x = t \Rightarrow {e^t} = x \Rightarrow {e^t}dt = dx,$ then
$I = \int_{}^{} {t(t + 2){e^t}dt} = {t^2}.\,{e^t} + c = x{(\log x)^2} + c.$
View full question & answer→MCQ 1191 Mark
$\int_{}^{} {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]dx = } $
AnswerCorrect option: B. $\frac{x}{{\log x}} + c$
b
(b)$\int_{}^{} {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx = \int_{}^{} {\frac{1}{{\log x}}\,dx - \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}\,dx} } $
$ = \frac{x}{{\log x}} + \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}\,.\,\frac{1}{x}x\,dx} - \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}} dx + c = \frac{x}{{\log x}} + c$.
View full question & answer→MCQ 1201 Mark
$\int_{}^{} {\left( {\frac{{2 + \sin 2x}}{{1 + \cos 2x}}} \right)\,\,{e^x}dx = } $
- A
${e^x}\cot x + c$
- B
$ - {e^x}\cot x + c$
- C
$ - {e^x}\tan x + c$
- ✓
${e^x}\tan x + c$
AnswerCorrect option: D. ${e^x}\tan x + c$
d
(d)$\int_{}^{} {\left( {\frac{{2 + \sin 2x}}{{1 + \cos 2x}}} \right){\rm{ }}{e^x}\,dx} = \int_{}^{} {\left( {\frac{{2{e^x}}}{{1 + \cos 2x}}} \right)dx} + \int_{}^{} {\frac{{{e^x}\sin 2x}}{{1 + \cos 2x}}dx} $
$ = \int_{}^{} {{e^x}{{\sec }^2}x\,dx} + \int_{}^{} {{e^x}\tan x\,dx = {e^x}\tan x + c} $.
View full question & answer→MCQ 1211 Mark
$\int {\cos ({{\log }_e}x)\,dx} $ is equal to
- ✓
$\frac{1}{2}x\{ \cos ({\log _e}x) + \sin ({\log _e}x)\} $
- B
$x\{ \cos ({\log _e}x) + \sin ({\log _e}x)\} $
- C
$\frac{1}{2}x\{ \cos ({\log _e}x) - \sin ({\log _e}x)\} $
- D
$x\{ \cos ({\log _e}x) - \sin ({\log _e}x)\} $
AnswerCorrect option: A. $\frac{1}{2}x\{ \cos ({\log _e}x) + \sin ({\log _e}x)\} $
a
(a) Let $I = \int {\cos ({{\log }_e}x)\,dx} $$ = \int {\cos ({{\log }_e}x)\,.\,1\,dx} $
$I = \cos ({\log _e}x).\,x - \int {\frac{{ - \sin ({{\log }_e}x)}}{x}} .\,x\,\,dx$
$ = x\cos ({\log _e}x) + \int {\sin ({{\log }_e}x)} \,\,dx$
$ = x\cos \,({\log _e}x) + \int {\sin \,({{\log }_e}x)} \,\,1\,\,dx$
$ = x\cos ({\log _e}x) + \sin ({\log _e}x).\,x - \int {\frac{{\cos ({{\log }_e}x)}}{x}x\,dx} $
$ = x\,\cos ({\log _e}x) + x\sin ({\log _e}x) - I$
$ \Rightarrow 2I = x\,[\cos \,({\log _e}x) + \sin \,({\log _e}x)]$
$ \Rightarrow I = \frac{x}{2}\,[\cos \,\,({\log _e}x) + \sin \,({\log _e}x)]$.
View full question & answer→MCQ 1221 Mark
$\int_{}^{} {\frac{{x{e^x}}}{{{{(1 + x)}^2}}}dx = } $
- A
$\frac{{{e^{ - x}}}}{{1 + x}} + c$
- B
$ - \frac{{{e^{ - x}}}}{{1 + x}} + c$
- ✓
$\frac{{{e^x}}}{{1 + x}} + c$
- D
$ - \frac{{{e^x}}}{{1 + x}} + c$
AnswerCorrect option: C. $\frac{{{e^x}}}{{1 + x}} + c$
c
(c)$\int_{}^{} {\frac{{x{e^x}}}{{{{(1 + x)}^2}}}\,dx = \int_{}^{} {\frac{{(x + 1 - 1)}}{{{{(1 + x)}^2}}}{e^x}dx} } $
$ = \int_{}^{} {{e^x}\left( {\frac{1}{{1 + x}} - \frac{1}{{{{(1 + x)}^2}}}} \right)\,dx} = \frac{{{e^x}}}{{1 + x}} + c$.
View full question & answer→MCQ 1231 Mark
$\int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)\,\,dx} $ is equal to
- A
$ - {e^x}\tan \,\left( {x/2} \right)$
- ✓
$ - {e^x}\cot \,\left( {x/2} \right)$
- C
$ - \frac{1}{2}{e^x}\tan \,\left( {\frac{x}{2}} \right)$
- D
$\frac{1}{2}{e^x}\cot \,\left( {\frac{x}{2}} \right)$
AnswerCorrect option: B. $ - {e^x}\cot \,\left( {x/2} \right)$
b
(b) $I = \int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx} $$ = \int {{e^x}\left( {\frac{{1 - \sin x}}{{2{{\sin }^2}(x2)}}} \right)\,dx} $
==> $I = \int {{e^x}\left( {\frac{1}{2}{\rm{cose}}{{\rm{c}}^2}\frac{x}{2} - \cot \frac{x}{2}} \right)} \,dx$
$\left( \because \,\,\int{{{e}^{x}}\left( f(x)+f'(x) \right)}={{e}^{x}}f(x)+c \right)$
$\therefore \,\,I = {e^x}\left( { - \cot \frac{x}{2}} \right) + c = - {e^x}\cot \frac{x}{2} + c$.
View full question & answer→MCQ 1241 Mark
$\int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}\,\,dx = \,\,} $
- A
$\frac{1}{{{{(x + 4)}^2}}} + c$
- B
$\frac{{{e^x}}}{{{{(x + 4)}^2}}} + c$
- ✓
$\frac{{{e^x}}}{{x + 4}} + c$
- D
$\frac{{{e^x}}}{{x + 3}} + c$
AnswerCorrect option: C. $\frac{{{e^x}}}{{x + 4}} + c$
c
(c) $I = \int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}dx} $$I = \int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}dx} $
$ \Rightarrow I = \int {{e^x}\,\left( {\frac{1}{{x + 4}} - \frac{1}{{{{(x + 4)}^2}}}} \right)\,dx} $
$\therefore I = \frac{{{e^x}}}{{x + 4}} + c$.
View full question & answer→MCQ 1251 Mark
$\int_{}^{} {{e^{{{\tan }^{ - 1}}x}}} \left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} \right)\;dx$ is equal to
- ✓
$x{e^{{{\tan }^{ - 1}}x}} + c$
- B
${x^2}{e^{{{\tan }^{ - 1}}x}} + c$
- C
$\frac{1}{x}{e^{{{\tan }^{ - 1}}x}} + c$
- D
AnswerCorrect option: A. $x{e^{{{\tan }^{ - 1}}x}} + c$
a
(a) Putting ${\tan ^{ - 1}}x = t$ and $\frac{{dx}}{{1 + {x^2}}} = dt,$ we get
$\int_{}^{} {{e^{{{\tan }^{ - 1}}x}}\left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} \right)} \,dx = \int_{}^{} {{e^t}(\tan t + {{\sec }^2}t)\,dt} $
$ = {e^t}\tan t + c = x\,{e^{{{\tan }^{ - 1}}x}} + c$
$\left[ {{\rm{Using }}\int_{}^{} {{e^x}\left\{ {f(x) + f'(x)} \right\}dx = {e^x}f(x) + C} } \right]$.
View full question & answer→MCQ 1261 Mark
$\int_{}^{} {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)\;dx = } $
- ✓
$\frac{1}{2}{e^{2x}}\cot 2x + c$
- B
$ - \frac{1}{2}{e^{2x}}\cot 2x + c$
- C
$ - 2{e^{2x}}\cot 2x + c$
- D
$2{e^{2x}}\cot 2x + c$
AnswerCorrect option: A. $\frac{1}{2}{e^{2x}}\cot 2x + c$
a
(a)$\int_{}^{} {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)} \,dx = \int_{}^{} {\frac{{{e^{2x}}\sin 4x}}{{1 - \cos 4x}}dx - 2\int_{}^{} {\frac{{{e^{2x}}}}{{1 - \cos 4x}}dx} } $ $ = \int_{}^{} {{e^{2x}}\cot 2x\,dx} - \int_{}^{} {{e^{2x}}{\rm{cose}}{{\rm{c}}^{\rm{2}}}2x\,dx} $ $ = \frac{{{e^{2x}}\cot 2x}}{2} + \int_{}^{} {2\frac{{{e^{2x}}}}{2}{\rm{cose}}{{\rm{c}}^{\rm{2}}}2x\,dx} - \int_{}^{} {{e^{2x}}{\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx} $ $ = \frac{1}{2}({e^{2x}}\cot 2x) + c.$
View full question & answer→MCQ 1271 Mark
The value of $\int {{e^{2x}}(2\sin 3x + 3\cos 3x)\,\,dx} $ is
- ✓
${e^{2x}}\sin 3x$
- B
${e^{2x}}\cos 3x$
- C
${e^{2x}}$
- D
${e^{2x}}(2\sin 3x)$
AnswerCorrect option: A. ${e^{2x}}\sin 3x$
a
(a) $\int {{e^{2x}}(2\sin 3x + 3\cos 3x)\,\,dx} $
$ = {e^{2x}}\sin 3x + c$.$\left\{ {\because \int {{e^{mx}}[mf(x) + f'(x)]\,dx = {e^{mx}}f(x) + c} } \right\}$
View full question & answer→MCQ 1281 Mark
$\int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx}$ is equal to
- A
${e^x}\tan \frac{x}{2} + C$
- B
${-e^x}\tan \frac{x}{2} + C$
- ✓
${-e^x}\cot \frac{x}{2} + C$
- D
${e^x}\cot \frac{x}{2} + C$
AnswerCorrect option: C. ${-e^x}\cot \frac{x}{2} + C$
c
$I=\int e^{x}\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2-\sin ^{2} \frac{x}{2}}\right) d x$
$ = \int {{{\rm{e}}^{\rm{x}}}} \left( {\frac{1}{2}\cos e{c^2}\frac{{\rm{x}}}{2} - \cot \frac{{\rm{x}}}{2}} \right){\rm{dx}}$
$ = - \int {{{\rm{e}}^{\rm{x}}}} \left( {\cot \frac{{\rm{x}}}{2} - \frac{1}{2}\cos e{c^2}\frac{{\rm{x}}}{2}} \right){\rm{dx}}$
$=-e^{x} \cdot \cot \frac{x}{2}+c$
View full question & answer→MCQ 1291 Mark
$\int {\frac{{{x^2}}}{{{x^2} + 4}}\,\,dx} $ equals to
- ✓
$x - 2{\tan ^{ - 1}}(x/2) + c$
- B
$x + 2{\tan ^{ - 1}}(x/2) + c$
- C
$x - 4{\tan ^{ - 1}}(x/2) + c$
- D
$x + 4{\tan ^{ - 1}}(x/2) + c$
AnswerCorrect option: A. $x - 2{\tan ^{ - 1}}(x/2) + c$
a
(a) $I = \int {\frac{{{x^2}}}{{{x^2} + 4}}dx} $$ = \int {\frac{{{x^2} + 4 - 4}}{{({x^2} + 4)}}dx} $
==> $I = \int {\left[ {1 - \frac{4}{{{x^2} + 4}}} \right]\,dx} $$ = \int {dx - \int {\frac{4}{{{x^2} + 4}}dx} } $
$ \Rightarrow I = x - 4\int {\frac{{dx}}{{{x^2} + {{(2)}^2}}}} $$ = x - \frac{4}{2}{\tan ^{ - 1}}(x/2) + c$
$ = x - 2{\tan ^{ - 1}}\frac{x}{2} + c$.
View full question & answer→MCQ 1301 Mark
$\int_{}^{} {\frac{1}{{\sqrt {1 - {e^{2x}}} }}\;dx = } $
- ✓
$x - \log [1 + \sqrt {1 - {e^{2x}}} ] + c$
- B
$x + \log [1 + \sqrt {1 - {e^{2x}}} ] + c$
- C
$\log [1 + \sqrt {1 - {e^{2x}}} ] - x + c$
- D
AnswerCorrect option: A. $x - \log [1 + \sqrt {1 - {e^{2x}}} ] + c$
a
(a)$\int_{}^{} {\frac{1}{{\sqrt {1 - {e^{2x}}} }}\,dx = \int_{}^{} {\frac{{{e^{ - x}}}}{{\sqrt {{e^{ - 2x}} - 1} }}\,dx} } $
Put ${e^{ - x}} = t \Rightarrow - {e^{ - x}}dx = dt,$ then it reduces to
$ - \int_{}^{} {\frac{1}{{\sqrt {{t^2} - 1} }}\,dt = - \log \left[ {t + \sqrt {{t^2} - 1} } \right] + c} $
$ = - \log \left[ {{e^{ - x}} + \sqrt {{e^{ - 2x}} - 1} } \right] = - \log \left[ {\frac{1}{{{e^x}}} + \frac{{\sqrt {1 - {e^{2x}}} }}{{{e^x}}}} \right]$
$ = - \log \left[ {1 + \sqrt {1 - {e^{2x}}} } \right] + \log {e^x} + c$
$ = x - \log \left[ {1 + \sqrt {1 - {e^{2x}}} } \right] + c.$
View full question & answer→MCQ 1311 Mark
$\int_{}^{} {\cos x\sqrt {4 - {{\sin }^2}x} } \;dx = $
- A
$\frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} - 2{\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c$
- ✓
$\frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} + 2{\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c$
- C
$\frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} + {\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c$
- D
AnswerCorrect option: B. $\frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} + 2{\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c$
b
(b) Putting $\sin x = t \Rightarrow \cos x\,dx = dt,$ we get
$\int_{}^{} {\cos x\sqrt {4 - {{\sin }^2}x\,} dx} = \int_{}^{} {\sqrt {4 - {t^2}} dt = \int_{}^{} {\sqrt {{{(2)}^2} - {t^2}} dt} } $
$ = \frac{t}{2}\sqrt {4 - {t^2}} + \frac{4}{2}{\sin ^{ - 1}}\frac{t}{2} + c$
$ = \frac{1}{2}\sin x\sqrt {4 - {{\sin }^2}x} + 2{\sin ^{ - 1}}\left( {\frac{1}{2}\sin x} \right) + c.$
View full question & answer→MCQ 1321 Mark
$\int_{}^{} {\frac{{{{\sec }^2}x\;dx}}{{\sqrt {{{\tan }^2}x + 4} }} = } $
- ✓
$\log \left[ {\tan x + \sqrt {{{\tan }^2}x + 4} } \right] + c$
- B
$\frac{1}{2}\log \left[ {\tan x + \sqrt {{{\tan }^2}x + 4} } \right] + c$
- C
$\log \left[ {\frac{1}{2}\tan x + \frac{1}{2}\sqrt {{{\tan }^2}x + 4} } \right] + c$
- D
AnswerCorrect option: A. $\log \left[ {\tan x + \sqrt {{{\tan }^2}x + 4} } \right] + c$
a
(a) Put $t = \tan x \Rightarrow dt = {\sec ^2}x\,dx,$ then
$\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{\sqrt {{{\tan }^2}x + 4} }}} = \int_{}^{} {\frac{1}{{\sqrt {{t^2} + {2^2}} }}} \,dt$
$ = \log [\tan x + \sqrt {{{\tan }^2}x + 4} ] + c.$
View full question & answer→MCQ 1331 Mark
The value of $\int_{}^{} {\frac{{dx}}{{x\sqrt {{x^4} - 1} }}} $ is
- ✓
$\frac{1}{2}{\sec ^{ - 1}}{x^2} + k$
- B
$\log x\sqrt {{x^4} - 1} + k$
- C
$x\log \sqrt {{x^4} - 1} + k$
- D
$\log \sqrt {{x^4} - 1} + k$
AnswerCorrect option: A. $\frac{1}{2}{\sec ^{ - 1}}{x^2} + k$
a
(a) $I = \int_{}^{} {\frac{{dx}}{{x\sqrt {{x^4} - 1} }}} $
Put ${x^2} = t \Rightarrow 2x\,dx = dt \Rightarrow dx = \frac{{dt}}{{2x}} = \frac{{dt}}{{2\sqrt t }}$
$\therefore \,\,\,I = \int_{}^{} {\frac{{dt}}{{2t\sqrt {{t^2} - 1} }}} = \frac{1}{2}{\sec ^{ - 1}}t + k = \frac{1}{2}{\sec ^{ - 1}}{x^2} + k$
View full question & answer→MCQ 1341 Mark
If $\int_{}^{} {\frac{1}{{(1 + x)\sqrt x }}\;dx = f(x) + A} $, where $ A$ is any arbitrary constant, then the function $f(x)$ is
AnswerCorrect option: B. $2{\tan ^{ - 1}}\sqrt x $
b
(b) $I = \int_{}^{} {\frac{{dx}}{{\sqrt x (1 + {{(\sqrt x )}^2})}}} $
Put $\sqrt x = t \Rightarrow \frac{1}{{2\sqrt x }}\,dx = dt$
$I = \int_{}^{} {\frac{{2\,dt}}{{1 + {t^2}}} = 2{{\tan }^{ - 1}}t + A} $
$\therefore \,\,\,I = 2{\tan ^{ - 1}}\sqrt x + A$; $\therefore \,\,\,f(x) = 2{\tan ^{ - 1}}\sqrt x $.
View full question & answer→MCQ 1351 Mark
$\int_{}^{} {\sqrt {{x^2} - 8x + 7} } \;dx = $
- A
$\frac{1}{2}(x - 4)\sqrt {{x^2} - 8x + 7} + 9\log [x - 4 + \sqrt {{x^2} - 8x + 7} ] + c$
- B
$\frac{1}{2}(x - 4)\sqrt {{x^2} - 8x + 7} - 3\sqrt 2 \log [x - 4 + \sqrt {{x^2} - 8x + 7} ] + c$
- ✓
$\frac{1}{2}(x - 4)\sqrt {{x^2} - 8x + 7} - \frac{9}{2}\log [x - 4 + \sqrt {{x^2} - 8x + 7} ] + c$
- D
AnswerCorrect option: C. $\frac{1}{2}(x - 4)\sqrt {{x^2} - 8x + 7} - \frac{9}{2}\log [x - 4 + \sqrt {{x^2} - 8x + 7} ] + c$
c
(c)$\int_{}^{} {\sqrt {{x^2} - 8x + 7} \,dx = \int_{}^{} {\sqrt {{{(x - 4)}^2} - {{(3)}^2}} \,dx} } $
Now apply formula of $\int_{}^{} {\sqrt {{x^2} - {a^2}} \,dx.} $
View full question & answer→MCQ 1361 Mark
$\int_{}^{} {\frac{{dx}}{{\sqrt {2x - {x^2}} }} = } $
- A
${\cos ^{ - 1}}(x - 1) + c$
- ✓
${\sin ^{ - 1}}(x - 1) + c$
- C
${\cos ^{ - 1}}(1 + x) + c$
- D
${\sin ^{ - 1}}(1 - x) + c$
AnswerCorrect option: B. ${\sin ^{ - 1}}(x - 1) + c$
b
(b)$\int_{}^{} {\frac{{dx}}{{\sqrt {2x - {x^2}} }}} = \int_{}^{} {\frac{{dx}}{{\sqrt {1 - {{(x - 1)}^2}} }}} = {\sin ^{ - 1}}(x - 1) + c.$
View full question & answer→MCQ 1371 Mark
$\int_{}^{} {\frac{{dx}}{{5 + 4\cos x}} = } $
- A
$\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan x} \right) + c$
- B
$\frac{1}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan x} \right) + c$
- ✓
$\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan \frac{x}{2}} \right) + c$
- D
$\frac{1}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan \frac{x}{2}} \right) + c$
AnswerCorrect option: C. $\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan \frac{x}{2}} \right) + c$
c
(c)$\int_{}^{} {\frac{{dx}}{{5 + 4\cos x}}} $$ = \int_{}^{} {\frac{{dx}}{{5 + 4\left[ {\frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}} \right]}}} = \int_{}^{} {\frac{{{{\sec }^2}\frac{x}{2}}}{{9 + {{\tan }^2}\frac{x}{2}}}} \,dx$
Put $\tan \frac{x}{2} = t,$ then it reduces to
$2\int_{}^{} {\frac{{dt}}{{{3^2} + {t^2}}} = \frac{2}{3}{{\tan }^{ - 1}}\left[ {\frac{1}{3}\tan \frac{x}{2}} \right] + c} $
Aliter : Apply direct formula
i.e., $\int_{}^{} {\frac{1}{{a + b\cos x}}\,dx} $, {a > b}
$ = \frac{2}{{\sqrt {{a^2} - {b^2}} }}{\tan ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{x}{2}} \right] + c$
We get $\int_{}^{} {\frac{{dx}}{{5 + 4\cos x}}} = \frac{2}{3}{\tan ^{ - 1}}\left\{ {\frac{1}{3}\tan \frac{x}{2}} \right\} + c.$
View full question & answer→MCQ 1381 Mark
$\int_{}^{} {\frac{1}{{1 + {{\cos }^2}x}}dx} = $
- A
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\tan x) + c$
- B
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{1}{2}\tan x} \right) + c$
- ✓
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}\tan x} \right) + c$
- D
AnswerCorrect option: C. $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}\tan x} \right) + c$
c
(c)$\int_{}^{} {\frac{{dx}}{{1 + {{\cos }^2}x}}} = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\sec }^2}x + 1}}} = \int_{}^{} {\frac{{{{\sec }^2}x}}{{{{\tan }^2}x + 2}}} \,dx$
$ = \int_{}^{} {\frac{{dt}}{{{t^2} + 2}} = \frac{1}{{\sqrt 2 }}{{\tan }^{ - 1}}\left( {\frac{t}{{\sqrt 2 }}} \right) + c} $ $\{$Putting $\tan x = t\} $
$ = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}\tan x} \right) + c$.
Trick : By inspection,
$\frac{d}{{dx}}\left\{ {\frac{1}{{\sqrt 2 }}{{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}\tan x} \right)} \right\} = \frac{1}{{\sqrt 2 }}\left( {\frac{1}{{1 + \frac{{{{\tan }^2}x}}{2}}}} \right)\frac{1}{{\sqrt 2 }}{\sec ^2}x$
$ = \frac{1}{2}\,.\,\frac{{2{{\sec }^2}x}}{{(2 + {{\tan }^2}x)}} = \frac{{{{\sec }^2}x}}{{1 + {{\sec }^2}x}} = \frac{1}{{1 + {{\cos }^2}x}}$.
View full question & answer→MCQ 1391 Mark
$\int_{}^{} {\frac{{dx}}{{1 + 3{{\sin }^2}x}} = } $
- A
$\frac{1}{3}{\tan ^{ - 1}}(3{\tan ^2}x) + c$
- ✓
$\frac{1}{2}{\tan ^{ - 1}}(2\tan x) + c$
- C
${\tan ^{ - 1}}(\tan x) + c$
- D
AnswerCorrect option: B. $\frac{1}{2}{\tan ^{ - 1}}(2\tan x) + c$
b
(b)$\int_{}^{} {\frac{{dx}}{{1 + 3{{\sin }^2}x}}} = \int_{}^{} {\frac{{dx}}{{{{\sin }^2}x + {{\cos }^2}x + 3{{\sin }^2}x}}} $
$ = \int_{}^{} {\frac{{dx}}{{4{{\sin }^2}x + {{\cos }^2}x}}} = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{4{{\tan }^2}x + 1}} = \frac{1}{4}\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\tan }^2}x + \frac{1}{4}}}} } $
Put $t = \tan x \Rightarrow dt = {\sec ^2}x\,dx,$ then it reduces to
$\frac{1}{4}\int_{}^{} {\frac{{dt}}{{{t^2} + {{\left( {\frac{1}{2}} \right)}^2}}}} = \frac{1}{4}2{\tan ^{ - 1}}(2t) + c$
$ = \frac{1}{2}{\tan ^{ - 1}}(2t) + c = \frac{1}{2}{\tan ^{ - 1}}(2\tan x) + c.$
View full question & answer→MCQ 1401 Mark
$\int_{}^{} {\frac{{dx}}{{2{x^2} + x + 1}}} \;$equals
- A
$\frac{1}{{\sqrt 7 }}{\tan ^{ - 1}}\left( {\frac{{4x + 1}}{{\sqrt 7 }}} \right) + c$
- B
$\frac{1}{{2\sqrt 7 }}{\tan ^{ - 1}}\left( {\frac{{4x + 1}}{{\sqrt 7 }}} \right) + c$
- C
$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{4x + 1}}{{\sqrt 7 }}} \right) + c$
- ✓
Answerd
(d)$I = \int_{}^{} {\frac{{dx}}{{2{x^2} + x + 1}}} = \int_{}^{} {\frac{{dx}}{{2\left( {{x^2} + \frac{x}{2} + \frac{1}{2}} \right)}}} $
$ = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{{x^2} + \frac{x}{2} + \frac{1}{{16}} - \frac{1}{{16}} + \frac{1}{2}}}} = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{{{\left( {x + \frac{1}{4}} \right)}^2} + {{\left( {\frac{{\sqrt 7 }}{4}} \right)}^2}}}} $
$ = \frac{1}{2}\frac{1}{{\frac{{\sqrt 7 }}{4}}}{\tan ^{ - 1}}\frac{{[x + (14)]}}{{\sqrt 7 4}}$$ = \frac{2}{{\sqrt 7 }}{\tan ^{ - 1}}\frac{{(4x + 1)}}{{\sqrt 7 }} + C$.
View full question & answer→MCQ 1411 Mark
$\int {\frac{{dx}}{{7 + 5\cos x}} = } $
- ✓
$\frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}\tan \frac{x}{2}} \right) + c$
- B
$\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
- C
$\frac{1}{4}{\tan ^{ - 1}}\left( {\tan \frac{x}{2}} \right) + c$
- D
$\frac{1}{7}{\tan ^{ - 1}}\left( {\tan \frac{x}{2}} \right) + c$
AnswerCorrect option: A. $\frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}\tan \frac{x}{2}} \right) + c$
a
(a)$I = \frac{{dx}}{{7 + 5\cos x}}$$ = \int {\frac{{dx}}{{7 + 5\,\left( {\frac{{1 - {{\tan }^2}(x/2)}}{{1 + {{\tan }^2}(x/2)}}} \right)}}} $
$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{7 + 7{{\tan }^2}(x/2) + 5 - 5{{\tan }^2}(x/2)}}} $
$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{12 + 2{{\tan }^2}(x/2)}}} $$ = \int {\frac{{\frac{1}{2}{{\sec }^2}(x/2)\,.dx}}{{6 + {{\tan }^2}(x/2)}}} $
$\tan \frac{x}{2} = t$ $⇒$ $\frac{1}{2}{\sec ^2}\frac{x}{2}dx = dt$
$I = \int {\frac{{dt}}{{{t^2} + ({{\sqrt {6)} }^2}}}} $ $ = \frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\frac{t}{{\sqrt 6 }} + c$
$ = \frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left| {\frac{{\tan (x/2)}}{{\sqrt 6 }}} \right| + c$
View full question & answer→MCQ 1421 Mark
$\int_{}^{} {\frac{{dx}}{{x[{{(\log x)}^2} + 4\log x - 1]}}} = $
- ✓
$\frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{\log x + 2 - \sqrt 5 }}{{\log x + 2 + \sqrt 5 }}} \right] + c$
- B
$\frac{1}{{\sqrt 5 }}\log \left[ {\frac{{\log x + 2 - \sqrt 5 }}{{\log x + 2 + \sqrt 5 }}} \right] + c$
- C
$\frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{\log x + 2 + \sqrt 5 }}{{\log x + 2 - \sqrt 5 }}} \right] + c$
- D
$\frac{1}{{\sqrt 5 }}\log \left[ {\frac{{\log x + 2 + \sqrt 5 }}{{\log x + 2 - \sqrt 5 }}} \right] + c$
AnswerCorrect option: A. $\frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{\log x + 2 - \sqrt 5 }}{{\log x + 2 + \sqrt 5 }}} \right] + c$
a
(a) Put $\log x = t \Rightarrow \frac{1}{x}\,dx = dt,$ then
$\int_{}^{} {\frac{{dx}}{{x[{{(\log x)}^2} + 4\log x - 1]}}} = \int_{}^{} {\frac{{dt}}{{{t^2} + 4t - 1}}} $
$ = \int_{}^{} {\frac{{dt}}{{{{(t + 2)}^2} - {{(\sqrt 5 )}^2}}} = \frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{t + 2 - \sqrt 5 }}{{t + 2 + \sqrt 5 }}} \right]} $
$ = \frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{\log x + 2 - \sqrt 5 }}{{\log x + 2 + \sqrt 5 }}} \right] + c$.
View full question & answer→MCQ 1431 Mark
$\int_{}^{} {\frac{{{x^3} - x - 2}}{{(1 - {x^2})}}\;dx = } $
- A
$\log \left( {\frac{{x + 1}}{{x - 1}}} \right) - \frac{{{x^2}}}{2} + c$
- B
$\log \left( {\frac{{x - 1}}{{x + 1}}} \right) + \frac{{{x^2}}}{2} + c$
- C
$\log \left( {\frac{{x + 1}}{{x - 1}}} \right) + \frac{{{x^2}}}{2} + c$
- ✓
$\log \left( {\frac{{x - 1}}{{x + 1}}} \right) - \frac{{{x^2}}}{2} + c$
AnswerCorrect option: D. $\log \left( {\frac{{x - 1}}{{x + 1}}} \right) - \frac{{{x^2}}}{2} + c$
d
(d)$\int_{}^{} {\frac{{{x^3} - x - 2}}{{(1 - {x^2})}}\,dx} = \int_{}^{} {\frac{{ - x(1 - {x^2})}}{{(1 - {x^2})}}\,dx - \int_{}^{} {\frac{2}{{1 - {x^2}}}\,dx} } $
$ = - \int_{}^{} {x\,dx} - 2\int_{}^{} {\frac{1}{{1 - {x^2}}}\,dx = \frac{{ - {x^2}}}{2} + \log \left( {\frac{{x - 1}}{{x + 1}}} \right) + c.} $
View full question & answer→MCQ 1441 Mark
$\int_{}^{} {\frac{{dx}}{{(1 + {x^2})\sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} }}} = $
- ✓
$\frac{1}{q}\log [q{\tan ^{ - 1}}x + \sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} ] + c$
- B
$\log [q{\tan ^{ - 1}}x + \sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} ] + c$
- C
$\frac{2}{{3q}}{({p^2} + {q^2}{\tan ^{ - 1}}x)^{3/2}} + c$
- D
AnswerCorrect option: A. $\frac{1}{q}\log [q{\tan ^{ - 1}}x + \sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} ] + c$
a
(a) Putting $q {\tan ^{ - 1}}x = t$
$ \Rightarrow \frac{q}{{1 + {x^2}}}dx = dt \Rightarrow \frac{1}{{1 + {x^2}}}dx = \frac{{dt}}{q}$
$ \Rightarrow \int_{}^{} {\frac{{dx}}{{(1 + {x^2})\sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} }}} = \frac{1}{q}\int_{}^{} {\frac{{dt}}{{\sqrt {{p^2} + {t^2}} }}} $
$ = \frac{1}{q}\log \left[ {q{{\tan }^{ - 1}}x + \sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} } \right] + c$.
View full question & answer→MCQ 1451 Mark
$\int_{}^{} {\frac{{{x^2}}}{{{{(9 - {x^2})}^{3/2}}}}\;dx = } $
- ✓
$\frac{x}{{\sqrt {9 - {x^2}} }} - {\sin ^{ - 1}}\frac{x}{3} + c$
- B
$\frac{x}{{\sqrt {9 - {x^2}} }} + {\sin ^{ - 1}}\frac{x}{3} + c$
- C
${\sin ^{ - 1}}\frac{x}{3} - \frac{x}{{\sqrt {9 - {x^2}} }} + c$
- D
AnswerCorrect option: A. $\frac{x}{{\sqrt {9 - {x^2}} }} - {\sin ^{ - 1}}\frac{x}{3} + c$
a
(a) Put $x = 3\sin \theta \Rightarrow dx = 3\cos \theta \,d\theta ,$
therefore $\int_{}^{} {\frac{{{x^2}}}{{{{(9 - {x^2})}^{32}}}}\,dx} = \int_{}^{} {\frac{{9{{\sin }^2}\theta }}{{{{(9 - 9{{\sin }^2}\theta )}^{32}}.\,3\cos \theta }}\,d\theta } $
$ = \int_{}^{} {\frac{{27{{\sin }^2}\theta \cos \theta }}{{27{{\cos }^3}\theta }}} \,d\theta = \int_{}^{} {{{\tan }^2}\theta \,d\theta } = \int_{}^{} {({{\sec }^2}\theta - 1)\,d\theta } $
$ = \tan \theta - \theta + c = \tan \left\{ {{{\sin }^{ - 1}}\left( {\frac{x}{3}} \right)} \right\} - {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + c$
$ = \tan {\tan ^{ - 1}}\left( {\frac{{\left( {\frac{x}{3}} \right)}}{{\sqrt {1 - ({x^2}9)} }}} \right) - {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + c$
$ = \frac{x}{{\sqrt {9 - {x^2}} }} - {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + c.$
View full question & answer→MCQ 1461 Mark
$\int_{}^{} {\frac{{dx}}{{4{{\sin }^2}x + 5{{\cos }^2}x}} = } $
- A
$\frac{1}{{\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c$
- B
$\frac{1}{{\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{\tan x}}{{\sqrt 5 }}} \right) + c$
- ✓
$\frac{1}{{2\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c$
- D
AnswerCorrect option: C. $\frac{1}{{2\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c$
c
(c)$\int_{}^{} {\frac{{dx}}{{4{{\sin }^2}x + 5{{\cos }^2}x}}} = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{4{{\tan }^2}x + 5}} = \frac{1}{4}\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\tan }^2}x + \frac{5}{4}}}} } $
Put $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt,$ then it reduces to
$\frac{1}{4}\int_{}^{} {\frac{{dt}}{{{t^2} + {{\left( {\frac{{\sqrt 5 }}{2}} \right)}^2}}} = \frac{2}{{4\sqrt 5 }}{{\tan }^{ - 1}}\left( {\frac{{2t}}{{\sqrt 5 }}} \right)} + c$
$ = \frac{1}{{2\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c.$
View full question & answer→MCQ 1471 Mark
The value of $\int {\frac{{\sqrt {({x^2} - {a^2})} }}{x}dx} $ will be
- ✓
$\sqrt {({x^2} - {a^2})} \, - a{\tan ^{ - 1}}\left[ {\frac{{\sqrt {({x^2} - {a^2})} }}{a}} \right]$
- B
$\sqrt {({x^2} - {a^2})} \, + a{\tan ^{ - 1}}\left[ {\frac{{\sqrt {({x^2} - {a^2})} }}{a}} \right]$
- C
$\sqrt {({x^2} - {a^2})} \, + {a^2}{\tan ^{ - 1}}[\sqrt {{x^2} - {a^2}} ]$
- D
${\tan ^{ - 1}}x/a + c$
AnswerCorrect option: A. $\sqrt {({x^2} - {a^2})} \, - a{\tan ^{ - 1}}\left[ {\frac{{\sqrt {({x^2} - {a^2})} }}{a}} \right]$
a
(a) Let $\sqrt {({x^2} - {a^2})} = t$ ==> ${x^2} - {a^2} = {t^2}$ ==> ${x^2} = {a^2} + {t^2}$
$\therefore$ $xdx = tdt$
$\int {\frac{{\sqrt {({x^2} - {a^2})} }}{x}dx} = \int {\frac{{\sqrt {({x^2} - {a^2})} \,x}}{{{x^2}}}dx} $
==> $I = \int {\frac{t}{{{a^2} + {t^2}}}tdt} $$ = \int {\frac{{{t^2}}}{{{a^2} + {t^2}}}dt} $
==> $I = \int {\left( {1 - \frac{{{a^2}}}{{{a^2} + {t^2}}}} \right)\,dt} $$ = t - {a^2}\frac{1}{a}{\tan ^{ - 1}}\left( {\frac{t}{a}} \right)$
==> $I = \sqrt {({x^2} - {a^2})} \, - a{\tan ^{ - 1}}\left[ {\frac{{\left\{ {\sqrt {({x^2} - {a^2})} } \right\}}}{a}} \right]$.
View full question & answer→MCQ 1481 Mark
$\int_{}^{} {x{{\sin }^{ - 1}}x\;dx} = $
- ✓
$\left( {\frac{{{x^2}}}{2} - \frac{1}{4}} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + c$
- B
$\left( {\frac{{{x^2}}}{2} + \frac{1}{4}} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + c$
- C
$\left( {\frac{{{x^2}}}{2} - \frac{1}{4}} \right){\sin ^{ - 1}}x - \frac{x}{4}\sqrt {1 - {x^2}} + c$
- D
$\left( {\frac{{{x^2}}}{2} + \frac{1}{4}} \right){\sin ^{ - 1}}x - \frac{x}{4}\sqrt {1 - {x^2}} + c$
AnswerCorrect option: A. $\left( {\frac{{{x^2}}}{2} - \frac{1}{4}} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + c$
a
(a)$\int_{}^{} {x{{\sin }^{ - 1}}xdx = \frac{{{x^2}}}{2}{{\sin }^{ - 1}}x - \int_{}^{} {\frac{1}{{\sqrt {1 - {x^2}} }}.\frac{{{x^2}}}{2}dx + c} } $
$ = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x - \frac{1}{2}\int_{}^{} { - \frac{{(1 - {x^2}) + 1}}{{\sqrt {1 - {x^2}} }}} dx + c$
$ = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{1}{2}\int_{}^{} {\sqrt {1 - {x^2}} dx - \frac{1}{2}\int_{}^{} {\frac{1}{{\sqrt {1 - {x^2}} }}dx + c} } $
$ = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + \frac{1}{4}{\sin ^{ - 1}}x - \frac{1}{2}{\sin ^{ - 1}}x + c$
$ = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} - \frac{1}{4}{\sin ^{ - 1}}x$
$ = \left( {\frac{{{x^2}}}{2} - \frac{1}{4}} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + c$.
View full question & answer→MCQ 1491 Mark
$\int_{}^{} {\frac{{dx}}{{2 + \cos x}} = } $
- A
$2{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
- ✓
$\frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
- C
$\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
- D
AnswerCorrect option: B. $\frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
b
(b)$\int_{}^{} {\frac{{dx}}{{2 + \cos x}} = \int_{}^{} {\frac{{dx}}{{2{{\sin }^2}\left( {\frac{x}{2}} \right) + 2{{\cos }^2}\left( {\frac{x}{2}} \right) + {{\cos }^2}\left( {\frac{x}{2}} \right) - {{\sin }^2}\left( {\frac{x}{2}} \right)}}} } $
$ = \int_{}^{} {\frac{{dx}}{{{{\sin }^2}\left( {\frac{x}{2}} \right) + 3{{\cos }^2}\left( {\frac{x}{2}} \right)}}} = \int_{}^{} {\frac{{{{\sec }^2}\left( {\frac{x}{2}} \right)}}{{{{\tan }^2}\left( {\frac{x}{2}} \right) + 3}}dx} $
Put $\tan \left( {\frac{x}{2}} \right) = t \Rightarrow {\sec ^2}\left( {\frac{x}{2}} \right)\,dx = 2dt,$ then it reduces to
$2\int_{}^{} {\frac{{dt}}{{{t^2} + 3}}} = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{t}{{\sqrt 3 }}} \right) + c = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{\tan \left( {\frac{x}{2}} \right)}}{{\sqrt 3 }}} \right) + c$.
View full question & answer→MCQ 1501 Mark
$\int_{}^{} {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}} \;dx = $
AnswerCorrect option: C. $\frac{{{e^{2x}}\tan x}}{2} + c$
c
(c)$\int_{}^{} {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}\,dx} = \int_{}^{} {{e^{2x}}\left[ {\frac{1}{{1 + \cos 2x}} + \frac{{\sin 2x}}{{1 + \cos 2x}}} \right]\,dx} $
$ = \int_{}^{} {{e^{2x}}\left[ {\frac{{{{\sec }^2}x}}{2} + \tan x} \right]} \,dx$
$ = \frac{1}{2}\int_{}^{} {{e^{2x}}{{\sec }^2}x\,dx} + \int_{}^{} {{e^{2x}}\tan x\,dx} $
$ = \frac{{{e^{2x}}\tan x}}{2} - \int_{}^{} {\frac{{{e^{2x}}{{\sec }^2}x}}{2}\,dx} + \int_{}^{} {\frac{{{e^{2x}}{{\sec }^2}x}}{2}\,dx} + c$
$ = \frac{{{e^{2x}}\tan x}}{2} + c.$
View full question & answer→