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Maths M.C.Q (1 Marks)

Integrals · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 60 questions.

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M.C.Q (1 Marks)

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MCQ 11 Mark
$\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$ is equal to :
  • A
    $\pi$
  • B
    Zero (0)
  • C
    $\int_0^{\pi / 2} \frac{2 \sin x}{1+\sin x \cos x} d x$
  • D
    $\frac{\pi^2}{4}$
Answer
Let $I=\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
$
\begin{array}{l}
\Rightarrow I=\int_0^{\pi / 2} \frac{\sin (\pi / 2-x)-\cos (\pi / 2-x)}{1+\sin (\pi / 2-x) \cos (\pi / 2-x)} d x \\
\Rightarrow I=\int_0^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \cdot \sin x} d x
\end{array}
$
Adding (i) and (ii), we get
$
2 I=\int_0^{\pi / 2} 0 d x=0 \Rightarrow I=0
$
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MCQ 21 Mark
If $\int_{-2}^3 x^2 d x=k \int_0^2 x^2 d x+\int_2^3 x^2 d x$, then the value of $k$ is
  • A
    2
  • B
    1
  • C
    $0$
  • D
    $\frac{1}{2}$
Answer
Given, $\int_{-2}^3 x^2 d x=k \int_0^2 x^2 d x+\int_2^3 x^2 d x$
$
\begin{array}{l}
\Rightarrow\left[\frac{x^3}{3}\right]_{-2}^3=k\left[\frac{x^3}{3}\right]_0^2+\left[\frac{x^3}{3}\right]_2^3 \\
\Rightarrow \frac{27}{3}+\frac{8}{3}=k\left(\frac{8}{3}\right)+\frac{27}{3}-\frac{8}{3} \Rightarrow \frac{8 k}{3}=\frac{16}{3} \Rightarrow k=2
\end{array}
$
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MCQ 31 Mark
The value of $\int_1^e \log x d x$ is
  • A
    $0$
  • B
    1
  • C
    $e$
  • D
    $e \log e$
Answer
Let $I=\int_1^e \log x d x$
Using integration by parts, we get $I=[x \log x]_1^e-\int_1^e \frac{1}{x} \cdot x d x$
$\begin{array}{l}I=[x \log x]_1^e-[x]_1^e=e \log e-\log 1-e+1 \\ =e-0-e+1=1 \quad[\because \log e=1 \text { and } \log 1=0]\end{array}$
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MCQ 41 Mark
The value of $\int_0^3 \frac{d x}{\sqrt{9-x^2}}$ is:
  • A
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{2}$
  • D
    $\frac{\pi}{18}$
Answer
We have, $\int_0^3 \frac{d x}{\sqrt{9-x^2}}=\left[\sin ^{-1} \frac{x}{3}\right]_0^3=\sin ^{-1} 1-\sin ^{-1} 0=\frac{\pi}{2}$
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MCQ 51 Mark
Which of these is equal to $\int e^{(x \log 5)} e^x d x$, where $C$ is the constant of integration?
  • ✓
    $\frac{(5 e)^x}{\log 5 e}+C$
  • B
    $\log 5^x+x+C$
  • C
    $5^x \varepsilon^x+C$
  • D
    $(5 e)^x \log x+C$
Answer
Correct option: A.
$\frac{(5 e)^x}{\log 5 e}+C$
$\frac{(5 e)^x}{\log 5 e}+C$
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MCQ 61 Mark
For any integer $n$, the value of $\int_{-\pi}^\pi e^{\cos ^2 x} \sin ^3(2 n+1) x d x$ is
  • A
    -1
  • B
    $0$
  • C
    1
  • D
    2
Answer
$\begin{array}{l}\text {Let } f(x)=e^{\cos ^2 x} \sin ^3(2 n+1) x \\ f(-x)=e^{\cos ^2(-x)} \sin ^3(2 n+1)(-x) \\ \quad=-e^{\cos ^2 x} \sin ^3(2 n+1) x=-f(x) \\ \therefore \quad \int_{-\pi}^\pi e^{\cos ^2 x} \sin ^3(2 n+1) x d x=0 \quad(\because f(-x)=-f(x))\end{array}$
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MCQ 71 Mark
If $\frac{d}{d x}(f(x))=\log x$, then $f(x)$ equals:
  • A
    $-\frac{1}{x}+C$
  • B
    $x(\log x-1)+C$
  • C
    $x(\log x+x)+C$
  • D
    $\frac{1}{x}+C$
Answer
We have, $\frac{d}{d x}(f(x))=\log x$
On integrating both sides, we get
$
\begin{array}{l}
\int \frac{d}{d x}(f(x))=\int 1 \cdot \log x d x \\
\Rightarrow f(x)=\log x \int 1 \cdot d x-\int\left(\frac{d}{d x}(\log x) \int 1 \cdot d x\right) d x \\
{[\text { Integrating by parts }]} \\
\Rightarrow f(x)=x \cdot \log x-\int \frac{1}{x} x x d x \Rightarrow f(x)=x \log x-x+C \\
\Rightarrow f(x)=x(\log x-1)+C
\end{array}$
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MCQ 81 Mark
$\int_{-1}^1 \frac{|x-2|}{x-2} d x, x \neq 2$ is equal to
  • A
    1
  • B
    -1
  • C
    2
  • D
    -2
Answer
Let $I=\int_{-1}^1 \frac{|x-2|}{x-2} d x$
$
=\int_{-1}^1 \frac{-(x-2)}{x-2} d x=\int_{-1}^1-1 \cdot d x=[-x]_{-1}^1=-[1-(-1)]=-2
$
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MCQ 91 Mark
$\int_0^{\frac{\pi}{6}} \sec ^2\left(x-\frac{\pi}{6}\right) d x$ is equal to :
  • A
    $\frac{1}{\sqrt{3}}$
  • B
    $-\frac{1}{\sqrt{3}}$
  • C
    $\sqrt{3}$
  • D
    $-\sqrt{3}$
Answer
$\begin{array}{l}\text {We have, } \int_0^{\pi / 6} \sec ^2\left(x-\frac{\pi}{6}\right) d x \\ =\left[\tan \left(x-\frac{\pi}{6}\right)\right]_0^{\pi / 6} \\ =\tan \left(\frac{\pi}{6}-\frac{\pi}{6}\right)-\tan \left(0-\frac{\pi}{6}\right) \\ =\tan 0^{\circ}-\tan \left(-\frac{\pi}{6}\right) \quad\left(\because \tan 0^{\circ}=0 \text { and } \tan (-\theta)=-\tan \theta\right) \\ =0+\tan (\pi / 6)=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\end{array}$
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MCQ 101 Mark
$\int_0^4\left(e^{2 x}+x\right) d x$ is equal to
  • A
    $\frac{15+e^8}{2}$
  • B
    $\frac{16-e^8}{2}$
  • C
    $\frac{e^8-15}{2}$
  • D
    $\frac{-e^8-15}{2}$
Answer
$\begin{array}{l}\text {Let } I=\int_0^4\left(e^{2 x}+x\right) d x=\left[\frac{e^{2 x}}{2}+\frac{x^2}{2}\right]_0^4 \\ =\frac{e^8}{2}+\frac{16}{2}-\frac{e^0}{2}-0=\frac{e^8}{2}+\frac{16}{2}-\frac{1}{2}=\frac{e^8+15}{2}\end{array}$
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MCQ 111 Mark
The value of $\int_0^{\pi / 6} \sin 3 x d x$ is:
  • A
    $-\frac{\sqrt{3}}{2}$
  • B
    $-\frac{1}{3}$
  • C
    $\frac{\sqrt{3}}{2}$
  • D
    $\frac{1}{3}$
Answer
$\begin{array}{l}\text {Let } I=\int_0^{\pi / 6} \sin 3 x d x \\ =\frac{-1}{3}[\cos 3 x]_0^{\pi / 6}=\frac{-1}{3}\left[\cos \frac{\pi}{2}-\cos 0\right]=\frac{-1}{3}(0-1)=\frac{1}{3}\end{array}$
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MCQ 121 Mark
\[\int \frac{\sec x}{\sec x-\tan x} d x \text { equals }\]
  • A
    $\sec x-\tan x+c$
  • B
    $\sec x+\tan x+c$
  • C
    $\tan x-\sec x+c$
  • D
    $-(\sec x+\tan x)+c$
Answer
$\begin{array}{l}\text {Let } I=\int \frac{\sec x}{\sec x-\tan x} d x \\ =\int \frac{\sec x(\sec x+\tan x)}{(\sec x-\tan x)(\sec x+\tan x)} d x=\int\left(\frac{\sec ^2 x+\sec x \tan x}{\sec ^2 x-\tan ^2 x}\right) d x \\ =\int \sec ^2 x d x+\int \sec x \tan x d x \quad\left[\because \sec ^2 x-\tan ^2 x=1\right] \\ =\tan x+\sec x+c\end{array}$
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MCQ 131 Mark
$\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x$ is equal to
  • A
    $\tan x-\cot x+C$
  • B
    $-\cot x-\tan x+C$
  • C
    $\cot x+\tan x+C$
  • D
    $\tan x-\cot x-C$
Answer
$\begin{array}{l}\text {Let, } I=\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x \\ =\int\left(\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cdot \cos ^2 x}\right) d x \quad \quad\left[\because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta\right] \\ =\int \frac{1}{\sin ^2 x} d x-\int \frac{1}{\cos ^2 x} d x=\int \operatorname{cosec}^2 x d x-\int \sec ^2 x d x \\ =-\cot x-\tan x+C\end{array}$
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MCQ 141 Mark
$\int e^{5 \log x} d x$ is equal to
  • A
    $\frac{x^5}{5}+C$
  • B
    $\frac{x^6}{6}+C$
  • C
    $5 x^4+C$
  • D
    $6 x^5+C$
Answer
$\begin{array}{l}\text {Let } I=\int e^{5 \log x} d x \\ =\int e^{\log x^5} d x=\int x^5 d x \quad\left[\because e^{\log x}=x\right] \\ =\frac{x^6}{6}+C\end{array}$
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MCQ 151 Mark
$\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x$ is equal to
  • A
    $\tan \left(x e^x\right)+c$
  • B
    $\cot \left(x e^x\right)+c$
  • C
    $\cot \left(e^x\right)+c$
  • D
    $\tan \left[e^x(1+x)\right]+c$
Answer
$\begin{array}{l}\text {Let } I=\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x \\ \text { Put } x e^x=t \Rightarrow\left(x e^x+e^x\right) d x=d t \Rightarrow e^x(x+1) d x=d t \\ \therefore \quad I=\int \frac{d t}{\cos ^2 t}=\int \sec ^2 t d t=\tan t+c=\tan \left(x e^x\right)+c\end{array}$
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MCQ 161 Mark
$\int_{-\pi / 4}^{\pi / 4} \sec ^2 x d x$ is equal to
  • A
    -1
  • B
    $0$
  • C
    1
  • D
    2
Answer
$\begin{array}{l}\text {Let } I=\int_{-\pi / 4}^{\pi / 4} \sec ^2 x d x=[\tan x]_{-\pi / 4}^{\pi / 4} \\ =\tan \frac{\pi}{4}-\tan \left(-\frac{\pi}{4}\right)=1+1=2\end{array}$
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MCQ 171 Mark
$\int_0^{\pi / 8} \tan ^2(2 x) d x$ is equal to
  • A
    $\frac{4-\pi}{8}$
  • B
    $\frac{4+\pi}{8}$
  • C
    $\frac{4-\pi}{4}$
  • D
    $\frac{4-\pi}{2}$
Answer
$\begin{array}{l}\text {Let } I=\int_0^{\pi / 8} \tan ^2(2 x) d x=\int_0^{\pi / 8}\left(\sec ^2(2 x)-1\right) d x \\ =\left(\frac{1}{2} \tan 2 x-x\right)_0^{\pi / 8}=\frac{1}{2} \tan 2\left(\frac{\pi}{8}\right)-\frac{\pi}{8}=\frac{1}{2} \tan \frac{\pi}{4}-\frac{\pi}{8} \\ =\frac{1}{2}-\frac{\pi}{8}=\frac{4-\pi}{8}\end{array}$
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MCQ 181 Mark
$\int \frac{e^x}{x+1}[1+(x+1) \log (x+1)] d x$ equals
  • A
    $\frac{e^x}{x+1}+c$
  • B
    $e^x \frac{x}{x+1}+c$
  • C
    $e^x \log (x+1)+e^x+c$
  • D
    $e^x \log (x+1)+c$
Answer
\[\begin{array}{l} 
\text { (d) : Let } I=\int \frac{e^x}{x+1}[1+(x+1) \log (x+1)] d x \\
=\int e^x\left[\frac{1}{x+1}+\log (x+1)\right] d x
\end{array}\]
It is of the form $\int e^x\left[f(x)+f^{\prime}(x) d x\right]$,
where $f(x)=\log (x+1)$ and $f^{\prime}(x)=\frac{1}{x+1}$
So, $I=e^x \log (x+1)+C$
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MCQ 191 Mark
$\int x^2 e^{x^3} d x$ equals
  • A
    $\frac{1}{3} e^{x^3}+C$
  • B
    $\frac{1}{3} e^{x^4}+C$
  • C
    $\frac{1}{2} e^{x^3}+C$
  • D
    $\frac{1}{2} e^{x^2}+C$
Answer
$\begin{array}{l}\text {Let } I=\int x^2 e^{x^3} d x \\ \text { Put } x^3=t \Rightarrow 3 x^2 d x=d t \\ \therefore \quad I=\int e^t \frac{d t}{3}=\frac{1}{3} e^t+C=\frac{1}{3} e^{x^3}+C\end{array}$
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MCQ 201 Mark
$\int e^x\left(\frac{x \log x+1}{x}\right) d x$ is equal to
  • A
    $\log \left(e^x \log x\right)+c$
  • B
    $\frac{e^x}{x}+c$
  • C
    $x \log x+e^x+c$
  • D
    $e^x \log x+c$
Answer
$\begin{array}{l}\text {Let } I=\int e^x\left(\log x+\frac{1}{x}\right) d x \\ \Rightarrow \quad I=e^x \log x+c \quad\left(\because \int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c\right)\end{array}$
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MCQ 211 Mark
Evaluate: $\int \frac{d x}{5-8 x-x^2}$
  • A
    $\frac{1}{\sqrt{21}} \log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C$
  • B
    $\frac{1}{2 \sqrt{21}} \log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C$
  • C
    $\frac{1}{\sqrt{21}} \log \left|\frac{\sqrt{21}-x-4}{\sqrt{21}+x+4}\right|+C$
  • D
    $\frac{1}{2 \sqrt{21}} \log \left|\frac{\sqrt{21}-x-4}{\sqrt{21}+x+4}\right|+C$
Answer
$
\begin{array}{l}
\text { (b) : Let } I=\int \frac{d x}{5-8 x-x^2}=\int \frac{d x}{21-(x+4)^2} \\
=\int \frac{d x}{(\sqrt{21})^2-(x+4)^2}=\frac{1}{2 \sqrt{21}} \log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C
\end{array}
$
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MCQ 221 Mark
Evaluate: $\int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x$
  • A
    $\frac{2}{\cos \frac{x}{2}+\sin \frac{x}{2}}+C$
  • B
    $\frac{-2}{\cos \frac{x}{2}-\sin \frac{x}{2}}+C$
  • ✓
    $\frac{-2}{\cos \frac{x}{2}+\sin \frac{x}{2}}+C$
  • D
    $\frac{2}{\cos \frac{x}{2}-\sin \frac{x}{2}}+C$
Answer
Correct option: C.
$\frac{-2}{\cos \frac{x}{2}+\sin \frac{x}{2}}+C$
(c): We have,
$
\begin{array}{l}
\quad \int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x=\int \frac{\cos ^2(x / 2)-\sin ^2(x / 2)}{\{\cos (x / 2)+\sin (x / 2)\}^3} d x \\
\text { Put } t=\cos \frac{x}{2}+\sin \frac{x}{2} \Rightarrow 2 d t=\left[\cos \frac{x}{2}-\sin \frac{x}{2}\right] d x \\
\Rightarrow \int \frac{\cos (x / 2)-\sin (x / 2)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2} d x=2 \int \frac{1}{t^2} d t \\
\quad=\frac{-2}{t}+C=\frac{-2}{\cos (x / 2)+\sin (x / 2)}+C
\end{array}
$
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MCQ 231 Mark
Evaluate: $\int \frac{x-4}{(x-2)^3} \cdot e^x d x$
  • A
    $\frac{e^x}{(x-2)^3}+C$
  • B
    $\frac{-e^x}{(x-2)^3}+C$
  • C
    $\frac{e^x}{(x-2)^2}+C$
  • D
    $\frac{-e^x}{(x-2)^2}+C$
Answer
$\begin{array}{l}\text { (c) }: \int \frac{x-4}{(x-2)^3} \cdot e^x d x=\int\left[\frac{x-2}{(x-2)^3}-\frac{2}{(x-2)^3}\right] e^x d x \\ =\int\left[\frac{1}{(x-2)^2}-\frac{2}{(x-2)^3}\right] e^x d x=\frac{e^x}{(x-2)^2}+C \\ {\left[\because \int\left[f(x)+f^{\prime}(x)\right] e^x d x=e^x f(x)+C\right]}\end{array}$
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MCQ 241 Mark
Evaluate: $\int_{-\pi}^\pi x^{10} \sin ^7 x d x$
  • A
    1
  • B
    2
  • C
    -1
  • ✓
    0
Answer
Correct option: D.
0
(d) : Let $I=\int_{-\pi}^\pi x^{10} \sin ^7 x d x$
Let $f(x)=x^{10} \sin ^7 x$
and $f(-x)=(-x)^{10}[\sin (-x)]^7=-x^{10} \sin ^7 x=-f(x)$
$\therefore f(x)$ is an odd function.
$\therefore \quad I=\int_{-\pi}^\pi x^{10} \sin ^7 x d x=0$
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MCQ 251 Mark
Evaluate: $\int \frac{\sin x}{1+\sin x} d x$
  • A
    $\sec x-\tan x+C$
  • B
    $\sec x+\tan x+x+C$
  • C
    $\sec x+\tan x+C$
  • ✓
    $\sec x-\tan x+x+C$
Answer
Correct option: D.
$\sec x-\tan x+x+C$
(d) : Let $I=\int \frac{\sin x}{1+\sin x} d x=\int \frac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)} d x$
$=\int \frac{\sin x-\sin ^2 x}{\cos ^2 x} d x=\int \sec x \tan x d x-\int \tan ^2 x d x$
$=\int \sec x \tan x d x-\int\left(\sec ^2 x-1\right) d x=\sec x-\tan x+x+C$
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MCQ 261 Mark
Evaluate : $\int \frac{x^3}{x+2} d x$
  • A
    $\frac{x^3}{3}-x^2-4 x-8 \log |x+2|+C$
  • ✓
    $\frac{x^3}{3}-x^2+4 x-8 \log |x+2|+C$
  • C
    $\frac{x^3}{3}+x^2+4 x+8 \log |x+2|+C$
  • D
    $\frac{x^3}{3}+x^2+4 x-8 \log |x+2|+C$
Answer
Correct option: B.
$\frac{x^3}{3}-x^2+4 x-8 \log |x+2|+C$
(b) : Let $I=\int \frac{x^3}{x+2} d x$
Dividing $x^3$ by $x+2$, we get
$
=\int\left(x^2-2 x+4-\frac{8}{x+2}\right) d x=\frac{x^3}{3}-x^2+4 x-8 \log |x+2|+C
$
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MCQ 271 Mark
Evaluate: $\int \sec ^2(7-4 x) d x$
  • A
    $\frac{1}{4} \tan (7-4 x)+C$
  • B
    $\frac{1}{4} \tan (7+4 x)+C$
  • C
    $\frac{-1}{4} \tan (7+4 x)+C$
  • ✓
    $\frac{-1}{4} \tan (7-4 x)+C$
Answer
Correct option: D.
$\frac{-1}{4} \tan (7-4 x)+C$
(d) : Let $I=\int \sec ^2(7-4 x) d x$
Put $7-4 x=t \Rightarrow d x=\frac{-1}{4} d t$
$
\therefore \quad I=\int \frac{\sec ^2 t}{-4} d t \Rightarrow I=\frac{\tan t}{-4}+C=\frac{\tan (7-4 x)}{-4}+C
$
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MCQ 281 Mark
Evaluate: $\int[\sin (\log x)+\cos (\log x)] d x$
  • ✓
    $x \sin (\log x)+C$
  • B
    $\sin (\log x)+C$
  • C
    $x \cos (\log x)+C$
  • D
    $\cos (\log x)+C$
Answer
Correct option: A.
$x \sin (\log x)+C$
(a) : Let $I=\int[\sin (\log x)+\cos (\log x)] d x$
Put $\log x=t \Rightarrow x=e^t \quad \Rightarrow d x=e^t d t$
$
\begin{aligned}
\therefore \quad I & =\int(\sin t+\cos t) e^t d t=e^t \sin t+C \\
& =x \sin (\log x)+C\left[\because\left[f(x)+f^{\prime}(x)\right] e^x d x=e^x f(x)+C\right)
\end{aligned}
$
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MCQ 291 Mark
Evaluate: $\int \frac{\left(a^x+b^x\right)^2}{a^x b^x} d x$
  • ✓
    $\frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log \frac{b}{a}}+2 x+C, a \neq b$
  • B
    $\frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log \frac{a}{b}}+2 x+C, a \neq b$
  • C
    $\left(\frac{a}{b}\right)^x+\left(\frac{b}{a}\right)^x+2 x+C, a \neq b$
  • D
    None of these
Answer
Correct option: A.
$\frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log \frac{b}{a}}+2 x+C, a \neq b$
(a) : We have, $\int \frac{\left(a^x+b^x\right)^2}{a^x b^x} d x=\int \frac{a^{2 x}+b^{2 x}+2 a^x b^x}{a^x b^x} d x$
$
=\int\left(\left(\frac{a}{b}\right)^x+\left(\frac{b}{a}\right)^x+2\right) d x=\frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log \frac{b}{a}}+2 x+C, a \neq b
$
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MCQ 301 Mark
Evaluate: $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$
  • A
    $57-5 \sqrt{5}$
  • B
    $\frac{57-\sqrt{5}}{5}$
  • C
    $\frac{57+5 \sqrt{5}}{5}$
  • ✓
    $\frac{57-5 \sqrt{5}}{5}$
Answer
Correct option: D.
$\frac{57-5 \sqrt{5}}{5}$
(d) : We have, $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$
Integrating by parts, we get
$
\begin{array}{l}
\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x=\left[\left(x^2+x\right) \cdot \sqrt{2 x+1}\right]_2^4-\int_2^4(2 x+1) \cdot \sqrt{2 x+1} d x \\
=(60-6 \sqrt{5})-\int_2^4(2 x+1)^{3 / 2} d x \\
=(60-6 \sqrt{5})-\frac{1}{5} \cdot\left[(2 x+1)^{5 / 2}\right]_2^4 \\
=(60-6 \sqrt{5})-\left(\frac{243}{5}-5 \sqrt{5}\right)=\left(\frac{57}{5}-\sqrt{5}\right)=\left(\frac{57-5 \sqrt{5}}{5}\right)
\end{array}
$
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MCQ 311 Mark
Evaluate: $\int \frac{\sec ^2 x}{2+\tan x} d x$
  • A
    $\log |\tan x|+C$
  • B
    $\log |2-\tan x|+C$
  • ✓
    $\log |2+\tan x|+C$
  • D
    none of these
Answer
Correct option: C.
$\log |2+\tan x|+C$
(c) : Let $I=\int \frac{\sec ^2 x}{2+\tan x} d x$
Put $2+\tan x=t \Rightarrow \sec ^2 x d x=d t$
$
\therefore \quad I=\int \frac{d t}{t}=\log |t|+C=\log |2+\tan x|+C
$
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MCQ 321 Mark
Evaluate: $\int \frac{d x}{\sqrt{1-2 x-x^2}}$
  • A
    $\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{1+x}{\sqrt{2}}\right)+C$
  • B
    $\frac{1}{\sqrt{2}} \log (1+x)+C$
  • C
    $\sin ^{-1}\left(\frac{1+x}{\sqrt{2}}\right)+C$
  • D
    $\frac{1}{\sqrt{2}} \log \left(\frac{1+x}{\sqrt{2}}\right)+C$
Answer
$
\begin{array}{l}
\text { (c) : Let } I=\int \frac{d x}{\sqrt{1-\left(x^2+2 x\right)}}=\int \frac{d x}{\sqrt{2-\left(x^2+2 x+1\right)}} \\
=\int \frac{d x}{\sqrt{2-(1+x)^2}}=\int \frac{d x}{\sqrt{(\sqrt{2})^2-(1+x)^2}}
\end{array}
$
Put $1+x=z \Rightarrow d x=d z$
$
\therefore I=\int \frac{d z}{\sqrt{(\sqrt{2})^2-z^2}}=\sin ^{-1} \frac{z}{\sqrt{2}}+C=\sin ^{-1}\left(\frac{1+x}{\sqrt{2}}\right)+C
$
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MCQ 331 Mark
Evaluate: $\int \frac{\cot x}{\sqrt[3]{\sin x}} d x$
  • A
    $\frac{-3}{\sqrt[3]{\sin x}}+C$
  • B
    $\frac{-2}{\sin ^3 x}+C$
  • C
    $\frac{3}{\sin ^{1 / 3} x}+C$
  • D
    None of these
Answer
$
\begin{array}{l}
\text { (a) : Let } I=\int \frac{\cot x}{\sqrt[3]{\sin x}} d x=\int \frac{\cos x}{\sin ^{1 / 3} x \cdot \sin x} d x \\
=\int \frac{\cos x}{\sin ^{4 / 3} x} d x=\int \sin ^{-4 / 3} x \cdot \cos x d x
\end{array}
$
Put $\sin x=t \Rightarrow \cos x d x=d t$
$
\Rightarrow I=\int t^{-4 / 3} d t=\frac{t^{-1 / 3}}{-1 / 3}+C=\frac{-3}{\sqrt[3]{\sin x}}+C
$
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MCQ 341 Mark
Evaluate : $\int_0^2 e^{3-4 x} d x$
  • A
    $\frac{-1}{4}\left(e^5-e^3\right)$
  • B
    $\frac{1}{4}\left(e^5-e^3\right)$
  • C
    $\frac{1}{4}\left(e^{-5}-e^3\right)$
  • D
    $\frac{-1}{4}\left(e^{-5}-e^3\right)$
Answer
$
\begin{array}{l}
\text { (d) : We have, } \int_0^2 e^{3-4 x} d x=\left[\frac{e^{3-4 x}}{-4}\right]_0^2 \\
=-\frac{1}{4}\left[e^{3-8}-e^{3-0}\right]=\frac{-1}{4}\left[e^{-5}-e^3\right]
\end{array}
$
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MCQ 351 Mark
Find the value of $\int_{-\pi / 2}^{\pi / 2}|\sin x| d x$.
  • A
    0
  • B
    1
  • C
    2
  • D
    3
Answer
(c) : $\because|\sin x|$ is an even function.$\therefore \quad \int_{-\pi / 2}^{\pi / 2}|\sin x| d x=2 \int_0^{\pi / 2}|\sin x| d x=2 \int_0^{\pi / 2} \sin x d x$ $=-2[\cos x]_0^{\pi / 2}=-2(0-1)=2$

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MCQ 361 Mark
Evaluate: $\int \frac{1}{\sin x+\sqrt{3} \cos x} d x$
  • ✓
    $\frac{1}{2} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right|+C$
  • B
    $\frac{1}{2} \log \left|\tan \frac{x}{2}\right|+C$
  • C
    $\frac{1}{2} \log \left|\tan \left(\frac{x}{2}-\frac{\pi}{6}\right)\right|+C$
  • D
    $\frac{1}{2} \log \left|\tan \left(x-\frac{\pi}{6}\right)\right|+C$
Answer
Correct option: A.
$\frac{1}{2} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right|+C$
(a) : Let $I=\int \frac{1}{\sin x+\sqrt{3} \cos x} d x$
$
\begin{array}{l}
=\frac{1}{2} \int \frac{d x}{\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x} \\
\Rightarrow \quad I=\frac{1}{2} \int \frac{1}{\sin \left(x+\frac{\pi}{3}\right)} d x=\frac{1}{2} \int \operatorname{cosec}\left(x+\frac{\pi}{3}\right) d x \\
\Rightarrow \quad I=\frac{1}{2} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right|+C \\
{\left[\because \int \operatorname{cosec} x d x=\log \left|\tan \frac{x}{2}\right|+C\right]} \\
\end{array}
$
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MCQ 371 Mark
Evaluate: $\int_0^{\pi / 4} \tan ^3 x d x$
  • A
    $(1-\log 2)$
  • B
    $(1+\log 2)$
  • ✓
    $\frac{1}{2}(1-\log 2)$
  • D
    $\frac{1}{2}(1+\log 2)$
Answer
Correct option: C.
$\frac{1}{2}(1-\log 2)$
(c)
$\begin{aligned}
\text { : Let } I= & \int_0^{\pi / 4} \tan ^3 x d x=\int_0^{\pi / 4}\left(\sec ^2 x-1\right) \tan x d x \\
& =\int_0^{\pi / 4} \sec ^2 x \tan x d x-\int_0^{\pi / 4} \tan x d x
\end{aligned}
$
Put $\tan x=t$ in first integral $\Rightarrow \sec ^2 x d x=d t$
When
$
\begin{array}{l}
x=0 \quad \Rightarrow t=0 \\
\therefore x=\pi / 4 \Rightarrow t=1 \\
I=\int_0^1 t d t-\int_0^{\pi / 4} \tan x d x=\left[\frac{t^2}{2}\right]_0^1-[\log |\sec x|]_0^{\pi / 4} \\
=\left(\frac{1}{2}-0\right)-\log \left|\sec \frac{\pi}{4}\right|+\log |\sec 0|=\frac{1}{2}(1-\log 2) \\
\end{array}
$
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MCQ 381 Mark
Evaluate: $\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x$
  • ✓
    $\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
  • B
    $\frac{-4}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
  • C
    $\frac{2}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
  • D
    $\frac{-2}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
Answer
Correct option: A.
$\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+C$
(a) : Let $I=\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x$
$
\Rightarrow \quad I=\int \frac{x\left(1-\frac{1}{x^3}\right)^{\frac{1}{4}}}{x^5} d x=\int \frac{\left(1-\frac{1}{x^3}\right)^{\frac{1}{4}}}{x^4} d x
$
Put $1-\frac{1}{x^3}=t \Rightarrow \frac{3}{x^4} d x=d t$
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MCQ 391 Mark
Evaluate: $\int 2^{2^{2^x}} 2^{2^x} 2^x d x$
  • ✓
    $\frac{1}{(\log 2)^3} 2^{2^{2^x}}+C$
  • B
    $\frac{1}{(\log 2)^3} 2^{2^x}+C$
  • C
    $\frac{1}{(\log 2)^2} 2^{2^x}+C$
  • D
    $\frac{1}{(\log 2)^4} 2^{2^{2^x}}+C$
Answer
Correct option: A.
$\frac{1}{(\log 2)^3} 2^{2^{2^x}}+C$
(a) : Let $I=\int 2^{2^{2^x}} 2^{2^x} 2^x d x$
Let $2^{2^{2^x}}=t \Rightarrow 2^{2^{2^x}} 2^{2^x} 2^x(\log 2)^3 d x=d t$
$
\Rightarrow I=\int \frac{1}{(\log 2)^3} d t=\frac{1}{(\log 2)^3} t+C=\frac{1}{(\log 2)^3} 2^{2^{2^x}}+C
$
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MCQ 401 Mark
Using fundamental theorem of calculus, which of following integrals can be solved.
(i) $\int_0^1 x^3 d x$
(ii) $\int x e^x d x$
(iii) $\int_2^3 \frac{1}{x} d x$
  • A
    Only (i)
  • B
    Only (ii)
  • ✓
    Only (i) and (iii)
  • D
    All (i), (ii) and (iii)
Answer
Correct option: C.
Only (i) and (iii)
(c) : (i) $\int_0^1 x^3 d x=\left[\frac{x^4}{4}\right]_0^1=\frac{1}{4}$
(ii) $\int x e^x d x=x e^x-e^x+C$
(iii) $\int_2^3 \frac{1}{x} d x=[\log x]_2^3=\log \frac{3}{2}$
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MCQ 411 Mark
Evaluate: $\int_0^2(x-[x]) d x$
  • A
    0
  • B
    -1
  • C
    1
  • D
    2
Answer
$
\begin{array}{l}
\text { (c) : Let } I=\int_0^2(x-[x]) d x=\int_0^2 x d x-\int_0^2[x] d x \\
=\left[\frac{x^2}{2}\right]_0^2-\int_0^1[x] d x-\int_1^2[x] d x=\frac{4}{2}-\int_0^1 0 d x-\int_1^2 1 d x \\
=2-0-[x]_1^2=2-[2-1]=2-1=1
\end{array}
$
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MCQ 421 Mark
Which of these is equal to $\int 2^{(x+3)} d x$, where $C$ is the constant of integration?
  • A
    $\frac{2^x}{\log 2}+C$
  • B
    $\frac{2^3}{\log 2}+C$
  • ✓
    $\frac{2^{(x+3)}}{\log 2}+C$
  • D
    $\frac{2^{(x-3)}}{\log 2}+C$
Answer
Correct option: C.
$\frac{2^{(x+3)}}{\log 2}+C$
(c): $\int 2^{(x+3)} d x=\int 2^x \cdot 2^3 d x=8 \int 2^x d x$
$=8 \cdot \frac{2^x}{\log 2}+C=\frac{2^{(x+3)}}{\log 2}+C$
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MCQ 431 Mark
Which of these is equal to $\int x e^{x^2} d x$, where $C$ is the constant of integration?
  • A
    $-\frac{e^{x^2}}{2}+C$
  • ✓
    $\frac{e^{x^2}}{2}+C$
  • C
    $\frac{e^x}{2}+C$
  • D
    $-\frac{e^x}{2}+C$
Answer
Correct option: B.
$\frac{e^{x^2}}{2}+C$
(b) : Let $I=\int x e^{x^2} d x$
Put $x^2=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{d t}{2}$
$\therefore \quad I=\frac{1}{2} \int e^t d t=\frac{e^t}{2}+C=\frac{e^{t^2}}{2}+C$
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MCQ 441 Mark
Evaluate : $\int_2^4 \frac{x}{x^2+1} d x$
  • ✓
    $\frac{1}{2} \log \left(\frac{17}{5}\right)$
  • B
    $\frac{1}{2} \log \left(\frac{5}{17}\right)$
  • C
    $\log \left(\frac{17}{5}\right)$
  • D
    $\log \left(\frac{5}{17}\right)$
Answer
Correct option: A.
$\frac{1}{2} \log \left(\frac{17}{5}\right)$
(a) : Let $I=\int_2^4 \frac{x}{x^2+1} d x$
Put $x^2+1=t \Rightarrow 2 x d x=d t \quad \Rightarrow \quad x d x=\frac{1}{2} d t$
Also, $x=2 \Rightarrow t=5$ and $x=4 \Rightarrow t=17$
$
\begin{array}{l}
\therefore \quad I=\frac{1}{2} \int_5^{17} \frac{d t}{t}=\frac{1}{2}[\log t]_5^{17} \\
=\frac{1}{2}[\log 17-\log 5]=\frac{1}{2} \log \left(\frac{17}{5}\right)
\end{array}
$
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MCQ 451 Mark
Find the value of $\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x$.
  • A
    $\tan x-\cot x+C$
  • B
    $-\tan x+\cot x+C$
  • C
    $\tan x+\cot x+C$
  • D
    $-\tan x-\cot x+C$
Answer
$
\begin{array}{l}
\text { (c) : We have, } \int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x \\
=\int\left(\sec ^2 x-\operatorname{cosec}^2 x\right) d x=\tan x+\cot x+C
\end{array}
$
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MCQ 461 Mark
Evaluate: $\int\left(2^x+2^{-x}\right)^2 d x$
  • A
    $\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+C$
  • B
    $\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+2 x+C$
  • C
    $\frac{1}{2 \log 2}\left(2^{2 x}+2^{-2 x}\right)+2 x+C$
  • D
    $\frac{1}{2 \log 2}\left(2^{2 x}+2^{-2 x}\right)+C$
Answer
$\begin{array}{l}\text { (b) : We have, } \int\left(2^x+2^{-x}\right)^2 d x=\int\left(2^{2 x}+2^{-2 x}+2\right) d x \\ =\frac{2^{2 x}}{(\log 2) \times 2}+\frac{2^{-2 x}}{(\log 2)(-2)}+2 \cdot x+C \\ =\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+2 x+C\end{array}$
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MCQ 471 Mark
Evaluate: $\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x$
  • A
    $\log _e\left(10^x-x^{10}\right)+C$
  • ✓
    $\log _e\left(10^x+x^{10}\right)+C$
  • C
    $\log _e\left(10^x+x^9\right)+C$
  • D
    $\log _e\left(10^x-x^9\right)+C$
Answer
Correct option: B.
$\log _e\left(10^x+x^{10}\right)+C$
(b) : Let $I=\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x$ Put $10^x+x^{10}=t$
$
\begin{aligned}
\Rightarrow \quad & \left(10^x \log _e 10+10 x^9\right) d x=d t \\
\therefore \quad & I=\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x=\int \frac{d t}{t} \\
& \quad=\log _e t+C=\log _e\left(10^x+x^{10}\right)+C
\end{aligned}
$
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MCQ 481 Mark
Ramkali is trying to find the solution of the following definite integrals :
(i) $\int_0^{2 \pi} \frac{d x}{e^{\sin x}+1}$
(ii) $\int_0^1 x^2 d x$
(iii) $\int_0^1 e^x d x$
Which of the above integrals solved by using of definite integral properties?
  • ✓
    Only (i)
  • B
    Only (i) and (ii)
  • C
    Only (ii) and (iii)
  • D
    None of the above
Answer
Correct option: A.
Only (i)
(a) : (i) Let $I=\int_0^{2 \pi} \frac{d x}{e^{\sin x}+1}$ ....(1)
$\Rightarrow \quad I=\int_0^{2 \pi} \frac{d x}{e^{\sin (2 \pi-x)}+1} \quad\left(\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right)$
$\Rightarrow \quad I=\int_0^{2 \pi} \frac{d x}{e^{-\sin x}+1} \Rightarrow I=\int_0^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x$ ....(2)
Adding (1) and (2), we get
$
2 I=\int_0^{2 \pi} 1 \cdot d x=2 \pi \quad \therefore \quad I=\pi
$
(ii) $\int_0^1 x^2 d x=\left[\frac{x^3}{3}\right]_0^1=\frac{1}{3}$
(iii) $\int_0^1 e^x d x=\left[e^x\right]_0^1=e^1-1$
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MCQ 491 Mark
Evaluate: $\int \tan x \tan 2 x \tan 3 x d x$
  • A
    $\frac{1}{3} \log |\sec 3 x|-\log |\sec x|+C$
  • B
    $\log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|+C$
  • C
    $\log |\sec x|-\frac{1}{2} \log |\sec 3 x|+\frac{1}{2} \log |\sec 2 x|+C$
  • ✓
    $\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$
Answer
Correct option: D.
$\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$
(d) : Let $I=\int \tan x \tan 2 x \tan 3 x d x$
$
\begin{array}{l}
\text { Since, } \tan 3 x=\tan (2 x+x)=\frac{\tan 2 x+\tan x}{1-\tan x \tan 2 x} \\
\Rightarrow \quad \tan x \tan 2 x \tan 3 x=\tan 3 x-\tan 2 x-\tan x & ...(i)\\
\therefore \quad I=\int(\tan 3 x-\tan 2 x-\tan x) d x & (From(i))\\
=\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C
\end{array}
$
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MCQ 501 Mark
Evaluate: $\int \frac{x^3-x^2+x-1}{x-1} d x$
  • A
    $\frac{x^3}{3}+x+C$
  • B
    $x^3+x+C$
  • C
    $\frac{x^3}{3}+x^2+C$
  • D
    $\frac{x^3}{3}+C$
Answer
$\begin{array}{l}\text { (a) : Let } I=\int \frac{x^3-x^2+x-1}{x-1} d x \\ =\int \frac{x^2(x-1)+1(x-1)}{x-1}=\int \frac{\left(x^2+1\right)(x-1)}{x-1} d x \\ =\int\left(x^2+1\right) d x=\frac{1}{3} x^3+x+C\end{array}$
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M.C.Q (1 Marks) - Maths STD 12 Science Questions - Vidyadip