M.C.Q (1 Marks)

MCQ 511 Mark

Evaluate : $\int \frac{d x}{\sqrt{x^2-3 x+2}}$

A
$\log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+C$
B
$\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+C$
C
$\log \left|\left(x-\frac{3}{2}\right)-\sqrt{x^2-3 x+2}\right|+C$
D
$\log \left|\left(x+\frac{3}{2}\right)-\sqrt{x^2-3 x+2}\right|+C$

Answer

$\begin{array}{l}\text { (b) : We have, } \int \frac{d x}{\sqrt{x^2-3 x+2}}=\int \frac{d x}{\sqrt{\left(x^2-3 x+\frac{9}{4}\right)-\frac{1}{4}}} \\ =\int \frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}}=\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+C\end{array}$

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MCQ 521 Mark

Evaluate: $\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x$

A
$\frac{4-\pi}{2 \sqrt{2}}$
B
$\frac{4+\pi}{2 \sqrt{2}}$
✓
$\frac{4-\pi}{4 \sqrt{2}}$
D
None of these

Answer

Correct option: C.

$\frac{4-\pi}{4 \sqrt{2}}$

(c): Let $\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x$ Put $\tan ^{-1} x=\theta \Rightarrow x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$ When, $x=0 \Rightarrow \theta=0$ and $x=1 \Rightarrow \theta=\frac{\pi}{4}$ $I=\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x=\int_0^{\pi / 4} \frac{\theta \tan \theta}{\sec ^3 \theta} \sec ^2 \theta d \theta$$=\int_0^{\pi / 4} \theta \sin \theta d \theta=[-\theta \cos \theta]_0^{\pi / 4}-\int_0^{\pi / 4}(-\cos \theta) d \theta$[Integrating by parts]
$=[-\theta \cos \theta]_0^{\pi / 4}+[\sin \theta]_0^{\pi / 4}=\frac{4-\pi}{4 \sqrt{2}}$

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MCQ 531 Mark

Evaluate: $\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x$

A
$\frac{3}{2} \sin ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$
B
$\frac{2}{3} \sin ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$
C
$\frac{2}{3} \cos ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$
D
$\frac{3}{2} \cos ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$

Answer

$\begin{array}{l}\text { (b) : Let } I=\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x \\ \text { Put } x^{3 / 2}=t \Rightarrow \frac{3}{2} x^{1 / 2} d x=d t\end{array}$
$\begin{array}{l}\therefore \quad I=\frac{2}{3} \int \frac{d t}{\sqrt{a^3-t^2}}=\frac{2}{3} \int \frac{d t}{\sqrt{\left(a^{3 / 2}\right)^2-t^2}} \\ =\frac{2}{3}\left[\sin ^{-1}\left(\frac{t}{a^{3 / 2}}\right)\right]+C=\frac{2}{3}\left[\sin ^{-1}\left(\frac{x^{3 / 2}}{a^{3 / 2}}\right)\right]+C \\ =\frac{2}{3} \sin ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C\end{array}$

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MCQ 541 Mark

Evaluate: $\int \sin ^3 x \cos ^3 x d x$

A
$\frac{-1}{32}\left\{\frac{-3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$
B
$\frac{1}{32}\left\{\frac{-3}{2} \cos 6 x+\frac{1}{6} \cos 2 x\right\}+C$
C
$\frac{1}{32}\left\{\frac{-3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$
D
None of these

Answer

$\begin{array}{l}\text { (c) : Let } I=\int \sin ^3 x \cos ^3 x d x \\ \Rightarrow \quad I=\frac{1}{8} \int(2 \sin x \cos x)^3 d x \\ \Rightarrow \quad I=\frac{1}{8} \int \sin ^3 2 x d x \Rightarrow \quad I=\frac{1}{8} \int \frac{3 \sin 2 x-\sin 6 x}{4} d x \\ \Rightarrow \quad I=\frac{1}{32}\left\{-\frac{3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C\end{array}$

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MCQ 551 Mark

Which of these is equal to $\int_0^1\left\{e^x+\sin \frac{\pi x}{4}\right\} d x$ ?

A
$e+1+\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}$
B
$e-1-\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}$
C
$e+1-\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}$
D
$e-1+\frac{2 \sqrt{2}}{\pi}-\frac{4}{\pi}$

Answer

$\begin{array}{l}\text { (b) : We have, } \int_0^1\left\{e^x+\sin \frac{\pi x}{4}\right\} d x \\ =\left[e^x\right]_0^1+\frac{4}{\pi}\left[-\cos \frac{\pi}{4} x\right]_0^1=e-1-\frac{4}{\sqrt{2} \pi}+\frac{4}{\pi} \\ =e-1-\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}\end{array}$

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MCQ 561 Mark

Which of these is equal to $\int x^2(a x+b)^{-2} d x$, where $C$ is the constant of integration?

A
$\frac{1}{a^3}\left(a x+b-\frac{b^2}{a x+b}-2 b \log (a x+b)\right)+C$
B
$\frac{1}{a^3}\left(a x+b+\frac{b^2}{a x+b}-2 b \log (a x+b)\right)+C$
C
$\frac{1}{a^3}\left(a x+b+\frac{b^2}{a x+b}+2 b \log (a x+b)\right)+C$
D
$\frac{1}{a^3}\left(a x+b-\frac{b^2}{a x+b}+2 b \log (a x+b)\right)+C$

Answer

14. (a) : Let $I=\int \frac{x^2}{(a x+b)^2} d x$
Put $a x+b=t \Rightarrow d x=\frac{1}{a} d t$
$
\begin{aligned}
\therefore \quad I & =\frac{1}{a^3} \int \frac{(t-b)^2}{t^2} d t=\frac{1}{a^3} \int\left(1+\frac{b^2}{t^2}-\frac{2 b}{t}\right) d t \\
& =\frac{1}{a^3}\left(t-\frac{b^2}{t}-2 b \log t\right)+C \\
& =\frac{1}{a^3}\left(a x+b-\frac{b^2}{a x+b}-2 b \log (a x+b)\right)+C
\end{aligned}
$

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MCQ 571 Mark

Evaluate: $\int \sqrt{(x-3)(5-x)} d x$

A
$\frac{1}{2}(x-4) \sqrt{(x-3)(5-x)}+\frac{1}{2} \cos ^{-1}(x-4)+C$
B
$\frac{1}{2}(x-4) \sqrt{(x-3)(5-x)}+\frac{1}{2} \sin ^{-1}(x-4)+C$
C
$\frac{1}{2} \sqrt{(x-3)(5-x)}+\frac{1}{2} \sin ^{-1}(x-4)+C$
D
None of these

Answer

$\begin{array}{l}\text { (b) : Let } I=\int \sqrt{(x-3)(5-x)} d x=\int \sqrt{-x^2+8 x-15} d x \\ \Rightarrow \quad I=\int \sqrt{-\left\{x^2-8 x+16-16+15\right\}} d x \\ \Rightarrow \quad I=\int \sqrt{-\left\{(x-4)^2-1^2\right\}} d x=\int \sqrt{1^2-(x-4)^2} d x \\ \Rightarrow \quad I=\frac{1}{2}(x-4) \sqrt{(x-3)(5-x)}+\frac{1}{2} \sin ^{-1}\left(\frac{x-4}{1}\right)+C\end{array}$

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MCQ 581 Mark

Evaluate:$\int\left(3 \sin x-2 \cos x+4 \sec ^2 x-5 \operatorname{cosec}^2 x\right) d x$

A
$-3 \cos x-2 \sin x+4 \tan x+5 \cot x+C$
B
$3 \cos x+2 \sin x+4 \tan x+5 \cot x+C$
C
$-3 \cos x+2 \sin x-4 \tan x-5 \cot x+C$
D
$-3 \cos x-2 \sin x-4 \tan x-5 \cot x+C$

Answer

$\begin{array}{l}\text { (a) : Let } I=\int\left(3 \sin x-2 \cos x+4 \sec ^2 x-5 \operatorname{cosec}^2 x\right) d x \\ \Rightarrow \quad I=3 \int \sin x d x-2 \int \cos x d x+4 \int \sec ^2 x d x-5 \int \operatorname{cosec}^2 x d x \\ \Rightarrow \quad I=-3 \cos x-2 \sin x+4 \tan x+5 \cot x+C\end{array}$

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MCQ 591 Mark

Evaluate: $\int\left(5 x^3+2 x^{-5}-7 x+\frac{1}{\sqrt{x}}+\frac{5}{x}\right) d x$

A
$\frac{5 x^4}{4}-\frac{1}{2 x^4}-\frac{7 x^2}{2}+2 \sqrt{x}-5 \log |x|+C$
B
$\frac{5 x^4}{4}-\frac{1}{2 x^4}-\frac{7 x^2}{2}+2 \sqrt{x}+5 \log |x|+C$
C
$\frac{5 x^4}{4}+\frac{1}{2 x^4}+\frac{7 x^2}{2}+2 \sqrt{x}+5 \log |x|+C$
D
$\frac{5 x^4}{4}+\frac{1}{2 x^4}+\frac{7 x^2}{2}+2 \sqrt{x}-5 \log |x|+C$

Answer

$
\begin{array}{l}
\text { (b) : We have } \int\left(5 x^3+2 x^{-5}-7 x+\frac{1}{\sqrt{x}}+\frac{5}{x}\right) d x \\
=5 \int x^3 d x+2 \int x^{-5} d x-7 \int x d x+\int x^{-1 / 2} d x+5 \int \frac{1}{x} d x \\
=5 \cdot \frac{x^4}{4}+2 \cdot \frac{x^{-4}}{(-4)}-7 \cdot \frac{x^2}{2}+\frac{x^{1 / 2}}{(1 / 2)}+5 \log |x|+C \\
=\frac{5 x^4}{4}-\frac{1}{2 x^4}-\frac{7 x^2}{2}+2 \sqrt{x}+5 \log |x|+C
\end{array}
$

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MCQ 601 Mark

Evaluate: $\int\left(e^{x \log a}+e^{a \log x}+e^{a \log a}\right) d x$

A
$a^x \log a+\frac{x^{a+1}}{a+1}+\frac{a^a}{x}+C$
B
$a^x \log a+(a+1) x^{a+1}+a^a x+C$
C
$\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^a x+C$
D
None of these

Answer

$
\begin{array}{l}
\text { (c) : Let } I=\int\left(e^{x \log a}+e^{a \log x}+e^{a \log a}\right) d x \\
=\int\left(e^{\log a^x}+e^{\log x^a}+e^{\log a^a}\right) d x=\int\left(a^x+x^a+a^a\right) d x \\
=\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^a x+c \quad \quad\left[\because e^{\log y=y]}\right]
\end{array}
$

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Maths M.C.Q (1 Marks)