Question 12 Marks
Find:
$\int\frac{\text{dx}}{5 - \text{8x - x}^{2}}$
Answer$\int\frac{\text{dx}}{5 - \text{8x - x}^{2}} = \int\frac{\text{dx}}{(\sqrt{21)^{2} - (\text{x + 4)}^{2}}}$
$= \frac{1}{2\sqrt{21}} \log \bigg|\frac{\sqrt{21} + \text{(x + 4)}}{\sqrt{21} - \text{(x + 4)}}\bigg| + \text{c}$
View full question & answer→Question 22 Marks
Find:
$\int \frac{\text{dx}}{\sqrt{3 - \text{2x - x}^{2}}}$
Answer$\text{I} = \int \frac{\text{dx}}{\sqrt{(2)^{2} - \text{(x + 1)}^{2}}}$
$= \sin^{-1} \bigg(\frac{\text{x + 1}}{2}\bigg) + \text{c}$
View full question & answer→Question 32 Marks
Find $\int \frac{\text{d}x}{x^{2} + 4x + 8}$
Answer$\int \frac{\text{dx}}{\text{x}^{2} + \text{4x + 8}} = \int \frac{\text{dx}}{(\text {x + 2)}^{2} + (2)^{2}}$
$= \frac{1}{2} \tan^{-1} \frac{\text{x + 2}}{2} + \text{C}$
View full question & answer→Question 42 Marks
Evaluate: $\int\frac{\cos2\text{x}+2\sin^2\text{x}}{\cos^2\text{x}}\text{dx}$
Answer$\int\frac{\cos2\text{x}+2\sin^2\text{x}}{\cos^2\text{x}}$
$\int\frac{2\cos^2\text{x}-1+2(1-\cos^2\text{x})}{\cos^2\text{x}}\text{dx}$
$\int\Big(2-\frac{1}{\cos^2\text{x}}+\frac{2}{\cos^2\text{x}}-1\Big)\text{dx}$
$\int\Big(1+\frac{1}{\cos^2\text{x}}\Big)\text{dx}$
$\text{x}+\int\text{sec}^2\text{x}\text{dx}$
$\text{x}+\tan\text{x}+\text{C}$
View full question & answer→Question 52 Marks
Find: $\int\sin\text{x}.\log\cos\text{x}\text{dx}.$
Answer$\int\sin\text{x}.\log\cos\text{x}\text{dx}=$
$\log\cos\text{x}\int\sin\text{x}\text{dx}-\int\Big(\frac{\text{d}}{\text{dx}}(\log\cos\text{x})\int\sin\text{x}\text{dx}\Big)\text{dx}$
$=-\cos\text{x}\log\cos\text{x}-\int\Big(\frac{\sin\text{x}}{\cos\text{x}}\times\cos\text{x}\Big)\text{dx}=-\cos\text{x}\log\cos\text{x}+\cos\text{x}+\text{C}.$
View full question & answer→Question 62 Marks
Find: $\int\frac{\tan^2\text{x}\sec^2\text{x}}{1-\tan^6\text{x}}\text{dx}.$
AnswerLet $\text{I}=\int\frac{\tan^2\text{x}\sec^2\text{x}}{1-\tan^6\text{x}}\text{dx}$
Let $\tan^3\text{x}=\text{t}$
$3\tan^2\text{x}\sec^2\text{x}\ \text{dx}=\text{dt}$
$\Rightarrow\text{I}=\int\frac{1}{3}\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{3}\int\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{6}\text{ln}\Big|\frac{1+\text{t}}{1-\text{t}}\Big|+\text{c},$
$\text{I}=\frac{1}{6}\text{ln}\Big|\frac{1+\tan^3\text{x}}{1-\tan^3\text{x}}\Big|+\text{c}$
View full question & answer→Question 72 Marks
Find: $\int\sin^{-1}(2\text{x})\text{dx}.$
Answer$\int\sin^{-1}(2\text{x})\text{dx}$
Using ILATE rule
$\text{x}\sin^{-1}2\text{x}-\int\frac{2\text{x}}{\sqrt{1-4\text{x}^2}}\text{dx}$
$\text{x}\sin^{-1}2\text{x}+\frac{1}{4}\int\frac{-8\text{x}}{\sqrt{1-4\text{x}^2}}\text{dx}$
Taking $1-4\text{x}^2=\text{t}$
$\Rightarrow-8\text{x}\times\text{dx}=\text{dt}$
$\text{x}\sin^{-1}2\text{x}+\frac{1}{4}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$\text{x}\sin^{-1}2\text{x}+\frac{1}{4}\frac{2\text{t}^{\frac{1}{2}}}{1}+\text{C}$
$=\text{x}\sin^{-1}2\text{x}+\frac{\text{t}^{\frac{1}{2}}}{2}+\text{C}$
$=\text{x}\sin^{-1}2\text{x}+\frac{\sqrt{1-4\text{x}^2}}{2}+\text{C}$
View full question & answer→Question 82 Marks
Find: $\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{dx}.$
Answer$\text{I}=\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
So, $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4}}$
Or, $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+2^2}}$
Since, we know
$\int\frac{\text{dx}}{\sqrt{\text{x}^2+\text{a}^2}}=\text{ln}\ \Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}$
$\text{I}=\text{ln}\ \Big|\text{t}+\sqrt{\text{t}^2+4}\Big|+\text{C}$
i.e., $\text{I}=\text{ln}|\tan\text{x}+\sqrt{{\tan}^2\text{x}+4}|+\text{C}$
View full question & answer→Question 92 Marks
Evaluate the definite integral in Exercise: $\int\limits_{0}^{\frac{\pi}{2}}\cos2\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\text{n}}{2}}\cos2\text{x}\ \text{dx}$ $\int\cos2\text{x}\ \text{dx}=\bigg(\frac{\sin2\text{x}}{2}\bigg)=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}\bigg(\frac{\pi}{2}\bigg)-\text{F}(0)$
$=\frac{1}{2}\bigg[\sin2\bigg(\frac{\pi}{2}\bigg)-\sin0\bigg]$
$=\frac{1}{2}[\sin\pi-\sin0]$
$=\frac{1}{2}[0-0]=0$
View full question & answer→Question 102 Marks
Evaluate the following integrals: $\int\frac{\text{e}^\text{x}}{\sqrt{16-\text{e}^{2\text{x}}}}\text{ dx}$
Answer$\int\frac{\text{e}^\text{x}\text{dx}}{\sqrt{16-(\text{e}^{\text{x}})^2}}$ Let $\text{e}^\text{x}=\text{t}$ $\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{\text{e}^\text{x}\text{dx}}{\sqrt{16-(\text{e}^{\text{x}})^2}}$ $=\int\frac{\text{dt}}{\sqrt{16-\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{4^2-\text{t}^2}}$
$=\sin^{-1}\Big(\frac{\text{t}}{4}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\text{e}^\text{x}}{4}\Big)+\text{C}$
View full question & answer→Question 112 Marks
Evaluate the following integrals: $\int\frac{\sec^2\text{x}}{\sqrt{4+\tan^2\text{x}}}\text{ dx}$
AnswerLet $\tan\text{x}=\text{t}$ $\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$ $\Rightarrow\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{ dx}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+2^2}}$ $=\log\Big|\text{t}+\sqrt{\text{t}^2+4}\Big|+\text{C}$
$=\log\Big|\tan+\sqrt{\tan^2\text{x}+4}\Big|+\text{C}$
View full question & answer→Question 122 Marks
Write a value of $\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$
Let $\tan^{-1}\text{x}=\text{t}$
$\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{t}^3}{1}\text{dt}$
$=\frac{\text{t}^4}{4}+\text{C}$
$\text{I}=\frac{(\tan^{-1}\text{x})^4}{4}+\text{C}$
View full question & answer→Question 132 Marks
Write a value of $\int\frac{\sec^2\text{x}}{(5+\tan\text{x})^4}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sec^2\text{x}}{(5+\tan\text{x})^4}\text{ dx}$
Let $5+\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}^4}$
$=-\frac{1}{3\text{t}^3}+\text{C}$
$\text{I}=-\frac{1}{3(5+\tan\text{x})^3}+\text{C}$
View full question & answer→Question 142 Marks
Write a value of $\int\text{a}^{\text{x}}\text{e}^{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\text{a}^{\text{x}}\text{e}^{\text{x}}\text{ dx}$ $=\int(\text{a}\text{e})^{\text{x}}\text{ dx}$
$=\frac{(\text{a}\text{e})^{\text{x}}}{\log\text{ae}}+\text{C}$
$\therefore\ \text{I}=\frac{(\text{a}\text{e})^{\text{x}}}{\log\text{ae}}+\text{C}$
View full question & answer→Question 152 Marks
Evaluate the definite integral in Exercise: $\int\limits_{0}^{1}\frac{\text{dx}}{1+\text{x}^{2}}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{4}\frac{\text{dx}}{1+\text{x}^{2}}$
$\int\frac{\text{dx}}{1+\text{x}^{2}}=\tan^{-1}\text{x}=\text{F}\text{(x)}$ By second fundamental theorem of calculus, we obtain $\text{I}=\text{F}(1)-\text{F}(0)$
$=\tan^{-1}(1)-\tan^{-1}(0)$
$=\frac{\pi}{4}$
View full question & answer→Question 162 Marks
Evaluate the following integrals: $\int\text{e}^{\text{x}}\Big(\frac{\text{x}-1}{2\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^\text{x}\frac{1}{2\text{x}}\text{dx}-\int\text{e}^{\text{x}}\frac{1}{2\text{x}^2}\text{dx}$
Integration by parts
$=\frac{\text{e}^{\text{x}}}{2\text{x}}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\Big(\frac{1}{2\text{x}}\Big)\Big)\text{dx}-\int\frac{\text{e}^{\text{x}}}{2\text{x}^2}\text{dx}$
$=\frac{\text{e}^{\text{x}}}{2\text{x}}+\int\frac{\text{e}^{\text{x}}}{2\text{x}^2}\text{dx}-\int\frac{\text{e}^{\text{x}}}{2\text{x}^2}\text{dx}$
$=\frac{\text{e}^{\text{x}}}{2\text{x}}+\text{C}$
View full question & answer→Question 172 Marks
Write the primitive or anti-derivative of $\text{f(x)}=\sqrt{\text{x}}=\frac{1}{\sqrt{\text{x}}}$.
Answer$\text{f(x)}=\sqrt{\text{x}}=\frac{1}{\sqrt{\text{x}}}$
Integrating both sides:
$\int\text{f(x)}\text{dx}=\int\Big(\sqrt{\text{x}}=\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}\Big)\text{dx}$
$=\bigg[\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg]+\bigg[\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg]+\text{C}$
$=\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}^{\frac{1}{2}}+\text{C}$
View full question & answer→Question 182 Marks
Integrate the functions in Exercises:
$\frac{1}{\sqrt{9-25\text{x}^2}}$
Answer$\int\frac{1}{\sqrt{9-25\text{x}^2}}\text{ dx}=\int\frac{1}{\sqrt{(3)^2-(5\text{x})^2}}\text{ dx}$
$=\frac{\sin^{-1}\frac{5\text{x}}{3}}{5\rightarrow\text{Coeff. of x}}+\text{c}$$\ \ \ \ \ \ \ \bigg[\because\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}=\sin^{-1}\frac{\text{x}}{\text{a}}\bigg]$
$=\frac{1}{5}\sin^{-1}\bigg(\frac{5\text{x}}{3}\bigg)+\text{c}$
View full question & answer→Question 192 Marks
Evaluate:
$\int\sqrt{\frac{1-\cos\ 2\text{x}}{2}}\text{dx}$
Answer$\int\sqrt{\frac{1-\cos\ 2\text{x}}{2}}\text{dx}$
$\int\sqrt{\frac{2\sin^2\text{x}}{2}}\text{dx}\ \ [\therefore1-\cos2\text{x}=2\sin^2\text{x]}$
$=\int\sin\text{x dx}$
$=-\cos\text{x}+\text{c}$
View full question & answer→Question 202 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\sqrt{\text{x}}}\cos\big(\text{e}^{\sqrt{\text{x}}}\big)}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\text{e}^{\sqrt{\text{x}}}\cos\big(\text{e}^{\sqrt{\text{x}}}\big)}{\sqrt{\text{x}}}\text{ dx}$
Let $\text{e}^{\sqrt{\text{x}}}=\text{t}$
$\Rightarrow\text{e}^{\sqrt{\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{e}^{\sqrt{\text{x}}}}{\sqrt{\text{x}}}\text{ dx}=2\text{dt}$
Now, $\int\frac{\text{e}^{\sqrt{\text{x}}}\cos\big(\text{e}^{\sqrt{\text{x}}}\big)}{\sqrt{\text{x}}}\text{ dx}$
$=2\int\cos\text{t}\text{ dt}$
$=2\sin\text{t}+\text{C}$
$=2\sin\Big(\text{e}^\sqrt{\text{x}}\Big)+\text{C}$
View full question & answer→Question 212 Marks
Evaluate $\int\frac{2}{1-\cos2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2}{1-\cos2\text{x}}\text{ dx}$
$=\int\frac{2}{\sin^2\text{x}+\cos^2\text{x}-(\cos^2\text{x}-\sin^2\text{x})}\text{ dx}$
$=\int\frac{2}{2\sin^2\text{x}}\text{ dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{ dx}$
$=\int\text{cosec}^2\text{x}\text{ dx}$
$=-\cot\text{x}+\text{C}$
View full question & answer→Question 222 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\cos2\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\cos(2\text{x})}\text{dx}$ $\Big[\therefore\ 1-\cos\text{A}=2\sin^2\Big(\frac{\text{A}}{2}\Big)\Big]$
$=\int\frac{\text{dx}}{2\sin^2\text{x}}$
$=\frac{1}{2}\int\text{cosec}^2\text{x dx}$
$=\frac{1}{2}[-\cot\text{x}]+\text{C}$
$=-\frac{1}{2}\cot\text{x}+\text{C}$
View full question & answer→Question 232 Marks
Evalute the following integrals:
$\int\frac{1-\sin\text{x}}{\text{x}+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\sin\text{x}}{\text{x}+\cos\text{x}}\text{dx}$
Putting $\text{x}+\cos\text{x}=\text{t}$
$\Rightarrow1-\sin\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln t}+\text{C}$
$=\text{ln}|\text{x}+\cos\text{x}|+\text{C}\ \big[\because\text{t}=\text{x}+\cos\text{x}\big]$
View full question & answer→Question 242 Marks
Evaluate the following integrals:
$\int\cos^{-1}(\sin\text{x})\text{dx}$
Answer$\int\cos^{-1}(\sin\text{x})\text{dx}$
$=\int\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big)\text{dx}$ $\Big[\therefore\ \sin\text{x}=\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big]$
$=\int\Big(\frac{\pi}{2}-\text{x}\Big)\text{dx}$
$=\frac{\pi\text{x}}{2}-\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 252 Marks
Evaluate $\int\frac{\text{e}^{\tan^{-1\text{x}}}}{1+\text{x}^2}\text{ dx}$
AnswerLet $\tan^{-1}\text{x}=\text{t}$
$\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$
Let $\text{I}=\int\frac{\text{e}^{\tan^{-1\text{x}}}}{1+\text{x}^2}\text{ dx}$
$=\int\text{e}^{\text{t}}\text{dt}$
$=\text{e}^{\text{t}}+\text{C}$
$=\text{e}^{\tan^{-1}}\text{x}+\text{C}$
View full question & answer→Question 262 Marks
Evaluate:
$\int\frac{\cos2\text{x}+2\sin^2\text{x}}{\sin^2\text{x}}\text{dx}$
Answer$\int\Big(\frac{\cos2\text{x}+2\sin^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1-2\sin^2\text{x}+2\sin^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$
$=\int\text{cosec}^2\text{x}\text{ dx}$
$=-\cot\text{x}+\text{c}$
View full question & answer→Question 272 Marks
Evaluate the following integrals:
$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$
Answer$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$
$=\int\limits^{\text{e}^2}_\text{e}\frac{\frac{1}{\text{x}}}{\log\text{x}}\text{ dx}$
$=\log\big[(\log{\text{x}})\big]^{\text{e}^2}_\text{e}$ $\Big[\int\frac{\text{f}'(\text{x})}{\text{f(x)}}\text{ dx}=\log\text{f(x)}+\text{C}\Big]$
$=\log\big(\log\text{e}^2\big)-\log(\log\text{e})$
$=\log(2\log\text{e})-\log(\log\text{e})$
$=\log2-\log1$ $(\log\text{e}=1)$
$=\log2-0$
$=\log2$
View full question & answer→Question 282 Marks
Verify the following:
$\int\frac{2\text{x}+3}{\text{x}^2+3\text{x}}\text{dx}=\log\big|\text{x}^2+3\text{x}\big|+\text{c}$
AnswerLet $\text{I}=\int\frac{2\text{x}+3}{\text{x}^2+3\text{x}}\text{dx}$
Put $\text{x}^2+3\text{x}=\text{t}$
$\Rightarrow\ (2\text{x}+3)\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{\text{t}}\text{dt}=\log|\text{t}|+\text{C}$ $=\log\big|(\text{x}^2+3\text{x})\big|+\text{C}$
View full question & answer→Question 292 Marks
Integrate the function in Exercise:
$\text{x} \ \sec^2\text{x}$
AnswerLet $\text{I}=\int\text{x}\sec^2\text{x dx}$
Taking x as first function and sec2x as second function and integrating by parts, we obtain.
$\text{I}=\text{x}\int\sec^2\text{x dx}-\int\Big[\Big\{\frac{\text{d}}{\text{dx}}\text{x}\int\sec^2\text{x} \ \text{dx}\Big\}\text{dx}\Big]$
$=\text{x}\tan\text{x}-\int1.\tan\text{x dx}$
$=\text{x}\tan\text{x}+\text{log}|\cos\text{x}|+\text{C}$
View full question & answer→Question 302 Marks
$\int\sin^2\text{bx dx}$
Answer$\int\sin^2\text{bx dx}$
$=\int\Big[\frac{1-\cos2\text{bx}}{2}\Big]\text{dx}$ $\Big[\therefore\sin^2\text{x}=\frac{1-\cos2\text{x}}{2}\Big]$
$=\frac{1}{2}\int(1-\cos2\text{bx})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{bx}}{2\text{b}}\Big]+\text{C}$
$=\frac{\text{x}}{2}-\frac{\sin2\text{bx}}{4\text{b}}+\text{C}$
View full question & answer→Question 312 Marks
Evalute the following integrals:
$\int\frac{\text{x}+1}{\text{x}(\text{x}+\log\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\text{x}+1}{\text{x}(\text{x}+\log\text{x})}\text{dx}$
Putting $\text{x}+\log\text{x}=\text{t}$
$\Rightarrow1+\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{x}+1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{x}+\log\text{x}|+\text{C}$
View full question & answer→Question 322 Marks
Evaluate the following integral:
$\int\frac{1}{\text{a}^2-\text{b}^2\text{x}^2}\text{ dx}$
Answer$\int\frac{1}{\text{a}^2-\text{b}^2\text{x}^2}\text{ dx}$
$=\frac{1}{\text{b}^2}\int\frac{\text{dx}}{\text{a}^2-\text{b}^2\text{x}}$
$=\frac{1}{\text{b}^2}\times\frac{1}{2\frac{\text{a}}{\text{b}}}\log\Bigg|\frac{\frac{\text{a}}{\text{b}}+\text{x}}{\frac{\text{a}}{\text{b}}-\text{x}}\Bigg|+\text{C}$ $\Big[\therefore\int\frac{\text{dx}}{\text{a}^2-\text{x}^2}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{x}}{\text{a}-\text{x}}\Big|+\text{C}\Big]$
$=\frac{1}{2\text{ab}}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}\Big|+\text{C}$
View full question & answer→Question 332 Marks
Evaluate the following integral:
$\int\frac{1}{\sqrt{\text{a}^2-\text{b}^2\text{x}^2}}\text{ dx}$
Answer$\int\frac{1}{\sqrt{\text{a}^2-\text{b}^2\text{x}^2}}\text{ dx}$
$=\int\frac{\text{dx}}{\sqrt{\text{b}^2\Big(\frac{\text{a}^2}{\text{b}^2}-\text{x}^2}\Big)}$
$=\frac{1}{\text{b}}\int\frac{\text{dx}}{\sqrt{\big(\frac{\text{a}}{\text{b}}\big)^2-\text{x}^2}}$
$=\frac{1}{\text{b}}\sin^{-1}\Big(\frac{\text{xb}}{\text{a}}\Big)+\text{C}$
View full question & answer→Question 342 Marks
Integrate the functions in Exercises:
$\frac{3\text{x}^2}{\text{x}^6+1}$
Answer$\text{Let I}=\int\frac{3\text{x}^2}{\text{x}^6+1}\text{ dx}$
$=\int\frac{3\text{x}^2}{(\text{x}^3)^2+1}\text{ dx} \ \ \ \ \ ...\text{(i)}$
Putting$\ \ \ \text{x}^3=\text{t} \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ 3\text{x}^2=\frac{\text{dt}}{\text{dx}} \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ 3\text{x}^2\text{ dx}=\text{dt}$
$\therefore \ \ \ \ \ $From eq. (i),$\ \ \ \ \text{I}=\int\frac{\text{dt}}{\text{t}^2+1}=\frac{1}{1}\tan^{-1}\frac{\text{t}}{1}+\text{c}$
$=\tan^{-1}\text{x}^3+\text{c}$
View full question & answer→Question 352 Marks
Evaluate the following integrals:
$\int\Big(\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}\Big)^2\text{dx}$
Answer$\int\Big(\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}\Big)^2\text{dx}$
$=\int\Big(\text{x}+\frac{1}{\text{x}}-2\Big)\text{dx}$
$=\int\text{xdx}+\int\frac{\text{dx}}{\text{x}}-2\int\text{dx}$
$=\frac{\text{x}^2}{2}+\ln|\text{x}|-2\text{x}+\text{C}$
View full question & answer→Question 362 Marks
Evaluate the following integrals:
$\int\text{x}\text{ cosec}^2\text{x dx}$
AnswerLet $\text{I}=\int\text{x cosec}^2\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}\int\text{cosec}^2\text{x dx}-\int(\int\text{cosec}^2\text{x dx})\text{dx}$
$=-\text{x}\cot\text{x}+\int\cot\text{x dx}$
$=-\text{x}\cot\text{x}+\log|\sin\text{x}|+\text{C}$
View full question & answer→Question 372 Marks
Evaluate the following integrals:
$\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
Answer$\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$=\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$
$=\frac{1}{3}\Big[\tan^{-1}\frac{\text{x}}{3}\Big]^3_0$
$=\frac{1}{3}\big(\tan^{-1}1-\tan^{-1}0\big)$
$=\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$
$=\frac{\pi}{12}$
View full question & answer→Question 382 Marks
Integrate the following integrals:
$\int\sin\text{mx}\cos\text{nx dx m}\neq\text{n}$
Answer$\int\sin(\text{mx})\cos(\text{nx) dx}$
$=\frac{1}{2}\int2\sin(\text{mx})\cos(\text{nx})\text{dx}$
$=\frac{1}{2}\int[\sin(\text{mx}+\text{nx})+\sin(\text{mx}-\text{nx})]\text{dx}$ $[\therefore2\sin\text{A}\cos\text{B}=\sin(\text{A}+\text{B})+\sin(\text{A}-\text{B})]$
$=\frac{1}{2}\Big[-\frac{\cos(\text{m+n})\text{x}}{\text{m}+\text{n}}-\frac{\cos(\text{m}-\text{n})\text{x}}{\text{m}-\text{n}}\Big]+\text{C}$
View full question & answer→Question 392 Marks
Evaluate the following integrals:
$\int\limits^1_{-2}\frac{|\text{x}|}{\text{x}}\text{ dx}$
AnswerLet $\int\limits^1_{-2}\frac{|\text{x}|}{\text{x}}\text{ dx}$
We have,
$|\text{x}|=\begin{cases}\text{x},&0\leq\text{x}\leq1\\-\text{x},&-2\leq\text{x}<0\end{cases}$
$\therefore\ \frac{|\text{x}|}{\text{x}}=\begin{cases}1,&0\leq\text{x}\leq1\\-1,&-2\leq\text{x}<0\end{cases}$
Therefore,
$\text{I}=\int\limits^0_{-2}-1\text{ dx}+\int\limits^1_01\text{ dx}$
$=-\big[\text{x}\big]^0_{-2}+\big[\text{x}\big]_0^1$
$=0-2+1-0$
$=-1$
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Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}} \cos^2\text{x}\text{ dx}$
Answer$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}} \cos^2\text{x}\text{ dx}$
$= \int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\frac{1+\cos2\text{x}}{2}\text{dx}$
$=\frac{1}{2}\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}(1+\cos2\text{x})\text{dx}$
$=\frac{1}{2}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\frac{1}{2}\Big(\frac{\pi}{2}+0+\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{2}$
View full question & answer→Question 412 Marks
Evaluate the definite integral in Exercise:
$\int\limits_{0}^{\frac{\pi}{4}}\tan\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^\frac{\pi}{4}\tan\text{x}\ \text{dx}$ $\int\tan\text{x}\ \text{dx}=-\text{log}|\cos\text{x|}=\text{F}\text{(x)}$ By second fundamental theoram of calculus, we obtain
$\text{I}=\text{F}\bigg(\frac{\pi}{4}\bigg)-\text{F}(0)$ $=-\text{log}|\cos\frac{\pi}{4}|+\text{log}|\cos0|$ $=-\text{log}\big|\frac{1}{\sqrt{2}}\big|+\text{log}|1|$ $=\text{log}(2)^\frac{1}{2}$ $=\frac{1}{2}\text{log}2$
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Evaluate $\int\frac{1}{\text{x}^2+16}\text{ dx}$
AnswerSince, $\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}$
Thus, $\int\frac{1}{\text{x}^2+16}\text{ dx}=\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}}{4}\Big)+\text{C}$
View full question & answer→Question 432 Marks
Prove the following Exercise:
$\int^{1}_{-1}\text{x}^{17}\cos^{4}\text{x}\ \text{dx}=0$
Answer$\text{Let I}=\int^{1}\limits_{-1}\text{x}^{17}\cos^{4}\text{x dx}$
Also, let $\text{f(x)}=\text{x}^{17}\cos^{4}\text{x}$
$\Rightarrow\text{f}\ (-\text{x)}=(-\text{x)}^{17}\cos^{4}(-\text{x)}=-\text{x}^{17}\cos^{4}\text{x}=-\text{f (x)}$
Therefore, f(x) is an odd function.
it is know that if f(x) is an odd function, then $\int^{\text{a}}\limits_{-\text{a}}\text{f (x) dx}=0$
$\therefore\text{I}=\int^{1}\limits_{-1}\text{x}^{17}\cos^{4}\text{x dx}=0$
Hence, the given result is Proved.
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Evaluate the following integrals:
$\int\limits^2_0\big[\text{x}\big]\text{dx}$
Answerwe have,
$\text{I}=\int\limits^2_0\big[\text{x}\big]\text{dx}$
$=\int\limits^1_0\big[\text{x}\big]\text{dx}+\int\limits^2_1\big[\text{x}\big]\text{dx}$
$=\int\limits^1_00\text{ dx}+\int\limits^2_1(\text{1})\text{dx}$ $\begin{bmatrix}\because\big[\text{x}\big]=\begin{cases}0,&0\leq\text{x}<1\\1,&1\leq\text{x}<2\end{cases}\end{bmatrix}$
$=0+\big[\text{x}\big]^{2}_1$
$=2-1$
$=1$
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Evaluate $\int\frac{\text{x}^2+4\text{x}}{\text{x}^3+6\text{x}^2+5}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+4\text{x}}{\text{x}^3+6\text{x}^2+5}\text{ dx}$
Let $\text{x}^3+6\text{x}^2+5=\text{t}$
$(3\text{x}^2+12\text{x})\text{dx}=\text{dt}$
$3(\text{x}^2+4\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{3}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{3}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{3}\log\big|\text{x}^3+6\text{x}^2+5\big|+\text{C}$
View full question & answer→Question 462 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x}\tan\text{x}}{3\sec\text{x}+5}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sec\text{x}\tan\text{x}}{3\sec\text{x}+5}\text{dx}$ then,
Putting $\sec\text{x}=\text{t}$
$\Rightarrow\frac{\text{dt}}{\text{dx}}=\sec\text{x}\tan\text{x}$
$\Rightarrow\text{dt}=\sec\text{x}\tan\text{x dx}$
$\therefore\text{I}=\int\frac{\text{dt}}{3\text{t}+5}$
$=\frac{1}{3}\text{ln}|3\text{t}+\text{5}|+\text{C}$
$=\frac{1}{3}\text{ln}|3\sec\text{x}+5|+\text{C}\ \big[\because\text{t}=\sec\text{x}\big]$
View full question & answer→Question 472 Marks
Evaluate the following integrals:
$\int\bigg\{\text{x}^2+\text{e}^{\log\text{x}}+\Big(\frac{\text{e}}{2}\Big)^\text{x}\bigg\}\text{dx}$
Answer$\int\bigg\{\text{x}^2+\text{e}^{\log\text{x}}+\Big(\frac{\text{e}}{2}\Big)^\text{x}\bigg\}\text{dx}$
$=\int\text{x}^2\text{dx}+\int\text{e}^{\log\text{x}}\text{dx}+\int\Big(\frac{\text{e}}{2}\Big)^\text{x}\text{dx}$
$=\frac{\text{x}^3}{3}+\int\text{xdx}+\int\Big(\frac{\text{e}}{2}\Big)\text{dx}$
$=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{2}+\frac{1}{\log\big(\frac{\text{e}}{2}\big)}\times\Big(\frac{\text{e}}{2}\Big)^\text{x}+\text{C}$
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Evaluate the following integrals:
$\int\text{x}^{\frac{5}{4}}\text{dx}$
Answer$\int\text{x}^{\frac{5}{4}}\text{dx}=\frac{\text{x}^{\frac{5}{4}}+1}{\frac{5}{4}+1}+\text{c}$
$=\frac{\text{x}^{\frac{5+4}{4}}+\text{c}}{\frac{5+4}{4}}$
$=\frac{4\text{x}^{\frac{9}{4}}}{9}+\text{c}$
View full question & answer→Question 492 Marks
Evaluate the following integrals:
$\int\cot^{-1}\Big(\frac{\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
Answer$\int\cot^{-1}\Big(\frac{\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
$=\int\cot^{-1}\Big(\frac{2\sin\text{x}\cos\text{x}}{2\sin^2\text{x}}\Big)\text{dx}$ $\big[\therefore\ \sin2\text{x}=2\sin\text{x}\cos\text{x} \text{ & }1-\cos2\text{x}=2\sin^2\text{x}\big]$
$=\int\cot^{-1}(\cot\text{x})\text{dx}$
$=\int\text{x dx}$
$=\frac{\text{x}^2}{2}+\text{C}$
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Evaluate $\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Let $\sqrt{\text{x}}=\text{t}$
$\frac{\text{dx}}{2\sqrt{\text{x}}}=\text{dt}$
$\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$
Putting $\sqrt{\text{x}}=\text{t}$ and $\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$
$\therefore\ \text{I}=2\int\sec^2+\text{dt}$
$=2\tan\text{t}+\text{C}$
$=2\tan(\sqrt{\text{x}})+\text{C}$ $(\because\text{t}=\sqrt{\text{x}})$
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