M.C.Q (1 Marks)

Question 1511 Mark

$\int\frac{\text{dx}}{2\sqrt{\text{x}}(1+\text{x})}=$

$\frac{1}{2}\tan(\sqrt{\text{x}})+\text{c}$
$\tan^{-1}(\sqrt{\text{x}})+\text{c}$
$2\tan^{-1}(\sqrt{\text{x}})+\text{c}$
None of these

Answer

$\tan^{-1}(\sqrt{\text{x}})+\text{c}$

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Question 1521 Mark

Evaluate: $\int\sec^{\frac{4}{3}}\text{x}\text{cosec}^{\frac{8}{3}}\text{xdx}.$

$\frac{3}{5}\tan^{\frac{-5}{3}}\text{x}-3\tan^{\frac{1}{3}}\text{x}+\text{c}$
$-\frac{3}{5}\tan^{\frac{-5}{3}}\text{x}+3\tan^{\frac{1}{3}}+\text{c}$
$-\frac{3}{5}\tan^{\frac{-05}{3}}\text{x}-3\tan^{\frac{1}{3}}+\text{c}$
$\text{None of these}$

Answer

$-\frac{3}{5}\tan^{\frac{-5}{3}}\text{x}+3\tan^{\frac{1}{3}}+\text{c}$

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Question 1531 Mark

$\int1.\text{dx}=$

$\text{x}+\text{k}$
$1+\text{k}$
$\frac{\text{x}^2}{2}+\text{k}$
$\log\text{x}+\text{k}$

Answer

$\text{x}+\text{k}$

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Question 1541 Mark

If x > a, $\int\frac{\text{dx}}{\text{x}^2-\text{a}^2}=$

$\frac{2}{2\text{a}}\text{log }\frac{\text{x-a}}{\text{x+a}}+\text{k}$
$\frac{2}{2\text{a}}\text{log }\frac{\text{x+a}}{\text{x-a}}+\text{k}$
$\frac{1}{\text{a}}\text{log}(\text{x}^2-\text{a}^2)+\text{k}$
$\log(\text{x}+\sqrt{\text{x}^2-\text{a}^2}+\text{k})$

Answer

$\frac{2}{2\text{a}}\text{log }\frac{\text{x-a}}{\text{x+a}}+\text{k}$

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Question 1551 Mark

$\frac{\text{x}}{\text{dx}}\int\text{f(x)}\text{dx}$ is equal to:

f'(x)
f(x)
f'(x’)
f(x) + c

Answer

f(x)

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Question 1561 Mark

$\int\cos^{-1}(\frac{1}{\text{x}})\text{dx}$ equals:

$\text{x}\sec^{-1}\text{x}+\log|\text{x}+\sqrt{\text{x}^2-1}|+\text{c}$
$\text{x}\sec^{-1}\text{x}-\log|\text{x}+\sqrt{\text{x}^2-1}|+\text{c}$
$-\text{x}\sec^{-1}\text{x}-\log|\text{x}+\sqrt{\text{x}^2-1}|+\text{c}$
$\text{None of these}$

Answer

$\text{x}\sec^{-1}\text{x}-\log|\text{x}+\sqrt{\text{x}^2-1}|+\text{c}$

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Question 1571 Mark

Choose the correct option from given four options:
If $\frac{\text{x}^3\text{dx}}{\sqrt{1+\text{x}^2}}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C},$ then:

$\text{a}=\frac{1}{3},\text{b}=1$
$\text{a}=\frac{-1}{3},\text{b}=1$
$\text{a}=\frac{-1}{3},\text{b}=-1$
$\text{a}=\frac{1}{3},\text{b}=-1$

Answer

$\text{a}=\frac{1}{3},\text{b}=-1$

Solution:
Given $\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{dx}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C}$
$\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{dx}$
Put $1+\text{x}^2=\text{t}^2$
$\Rightarrow\ 2\text{x dx}=2\text{t dt}$
$\therefore\ \text{I}\int\frac{\text{t}(\text{t}^2-1)}{\text{t}}\text{dt}$ $=\frac{\text{t}^3}{3}-\text{t}+\text{C}=\frac{1}{3}(1+\text{x}^2)^{\frac{3}{2}}-\sqrt{1+\text{x}^2}+\text{C}$
$\therefore\ \text{a}=\frac{1}{3}$ and $\text{b}=-1$

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Question 1581 Mark

$\text{I}=\int\frac{(\text{x+a})^3}{\text{x}^3}\text{dx}$ is equal to:

$\text{x}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}-\frac{\text{a}^3}{2\text{x}^2}+\text{c}$
$\text{x}^{2}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}-\frac{\text{a}^3}{2\text{x}^2}+\text{c}$
$\text{x}^{3}+3\text{a}\log\text{x}-\frac{2\text{a}^2}{\text{x}}-\frac{\text{3a}^3}{2\text{x}^2}+\text{c}$
${1}+2\text{a}\log\text{x}-\frac{2\text{a}^2}{\text{x}}-\frac{\text{3a}^3}{2\text{x}^2}+\text{c}$

Answer

$\text{x}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}-\frac{\text{a}^3}{2\text{x}^2}+\text{c}$

Solution:
$\text{I}=\int\frac{(\text{x+a})^3}{\text{x}^3}\text{dx}$
$=\int\frac{\text{x}^3+\text{a}^3+3\text{ax}^2+3\text{a}^2\text{x}}{\text{x}^3}\text{dx}$
$=\int\Big(1+\frac{\text{a}^3}{\text{x}^3}+\frac{3\text{a}}{\text{x}}+\frac{3\text{a}^2}{\text{x}^2}\Big)\text{dx}$
$=\text{x}-\frac{\text{a}^3}{2\text{x}^2}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}+\text{c}$
$\text{x}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}-\frac{\text{a}^3}{2\text{x}^2}+\text{c}$

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Question 1591 Mark

If $\int\limits^\alpha_0\frac{1}{1+4\text{x}^2}\text{ dx}=\frac{\pi}{8},$ then a equals:

$\frac{\pi}{2}$
$\frac{1}{2}$
$\frac{\pi}{4}$
$1$

Answer

$\frac{1}{2}$

Solution:
$\int\limits^\alpha_0\frac{1}{1+4\text{x}^2}\text{ dx}=\frac{\pi}{8}$
$\Rightarrow\int\limits^\alpha_0\frac{1}{1+(2\text{x)}^2}\text{ dx}=\frac{\pi}{8}$
$\Rightarrow \frac{1}{2}\big[\tan^-2\text{x}\big]^\alpha_0=\frac{\pi}{8}$
$\Rightarrow \frac{1}{2}\tan^{-1}2\alpha=\frac{\pi}{8}$
$\Rightarrow 2\alpha=\tan\frac{\pi}{4}$
$\Rightarrow2\alpha=1$
$\therefore\alpha=\frac{1}{2}$

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Question 1601 Mark

Choose the correct option from given four options:
$\int\frac{\text{x}}{\text{x}+1}$ is equal to:

$\text{x}+\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$
$\text{x}+\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$
$\text{x}-\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}-\log|1+\text{x}|+\text{C}$
$\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1+\text{x}|+\text{C}$

Answer

$\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1+\text{x}|+\text{C}$

Solution:
Let $\text{I}=\int\frac{\text{x}}{\text{x}+1}\text{dx}$
We know that, $\frac{\text{x}^3}{\text{x}+1}$ is an improper fraction.
To convert it into proper fraction, we have to divide numerator by denominator.
After performing long division, we get
$\frac{\text{x}^3}{\text{x}+1}=(\text{x}^2-\text{x}+1)-\frac{1}{(\text{x}+1)}$
$\therefore\ \text{I}=\int\Big((\text{x}^2-\text{x}+1)-\frac{1}{(\text{x}+1)}\Big)\text{dx}$
$=\frac{\text{x}^3}{3}-\frac{\text{x}^2}{2}+\text{x}-\log|\text{x}+1|+\text{C}$

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Question 1611 Mark

$\int\limits^1_0\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}\text{dx}$ is equal to:

$0$
${\pi}$
$\frac{\pi}{2}$
$\frac{\pi}{4}$

Answer

$\frac{\pi}{2}$

Solution:
We have,
$\text{I}=\int\limits^1_0\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}\text{dx}$
We know since $\int\text{f}'(\text{x})=\text{f}(\text{x})$
$\text{f}(\text{x})=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}$
Therefore, $\text{I}=\Big[\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big]^1_0$
$=\sin^{-1}(1)-\sin^{-1}(0)$
$=\frac{\pi}{2}$

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Question 1621 Mark

$\int\text{x}\sec\text{x}^2\text{ dx}$ is equal to:

$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
$\frac{\text{x}^2}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
$2\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
none of these.

Answer

$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$

Solution:
$\text{I}=\int\text{x}\sec\text{x}^2\text{ dx}$
Put $\text{x}^2=\text{t}$
$=\text{x}=\sqrt{\text{t}}$
$2\text{xdx}=\text{dt}$
$\text{xdx}=\frac{\text{dt}}{2}$
$\text{I}=\int\sec\text{t}\frac{\text{dt}}{2}$
$\text{I}=\frac{1}{2}\log(\sec\text{t}+\tan\text{t})+\text{C}$
$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$

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Question 1631 Mark

$\int\frac{2\text{dx}}{\sqrt{1-4\text{x}^2}}=$

$\tan^{-1}(2\text{x})+\text{c}$
$\cot^{-1}(2\text{x})+\text{c}$
$\cos^{-1}(2\text{x})+\text{c}$
$\sin^{-1}(2\text{x})+\text{c}$

Answer

$\sin^{-1}(2\text{x})+\text{c}$

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Question 1641 Mark

The value of $\int\frac{1}{\text{x}+\text{x}\log\text{x}}\text{ dx}$ is:

$1+\log\text{x}$
$\text{x}+\log\text{x}$
$\text{x}\log\text{x}(1+\log\text{x})$
$\log(1+\log\text{x})$

Answer

$\log(1+\log\text{x})$

Solution:
$\text{I}=\int\frac{1}{\text{x}+\text{x}\log\text{x}}\text{ dx}$
$\text{I}=\int\frac{\text{dx}}{\text{x}+(1+\log\text{x})}$
Put $1+\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\text{t}}\text{ dt}$
$\text{I}=\log|\text{t}|+\text{C}$
$\text{I}=\log(1+\log\text{x})+\text{C}$

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Question 1651 Mark

$\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\log\tan\text{x dx}$ is equal to:

$\pi$
$\frac{\pi}{2}$
$0$
$2\pi$

Answer

Solution:
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\log\tan\text{x dx}\ ....(\text{i})$
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin(\pi-2\text{x})\log\tan\big(\frac{\pi}{2}-\text{x}\big)\text{dx}$
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\log\cot\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}+\log\cot\text{x}\big)\text{dx}$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}\cot\text{x}\big)\text{dx}$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin\text{x}(\log1)\text{dx}$
$\text{I}=0$

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Question 1661 Mark

$\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}$ is equal to:

$\frac{1}{5\text{x}}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{c}$
$\frac{1}{5}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{c}$
$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}}+4\Big)^{-5}+\text{c}$
$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{c}$

Answer

$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{c}$

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Question 1671 Mark

If $\int\frac{1}{(\text{x}+2)(\text{x}^2+1)}\text{ dx}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C},$ then

$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$
$\text{a}=\frac{1}{10},\text{ b}=\frac{2}{5}$
$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$
$\text{a}=\frac{1}{10},\text{ b}=\frac{2}{5}$

Answer

$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$

Solution:
$\text{I}=\int\frac{1}{(\text{x}+2)(\text{x}^2+1)}\text{ dx}$
Consider,
$\frac{1}{(\text{x}+2)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{(\text{x}^2+1)}$
$1=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+2)$
Comaring coefficeints and solving it simultaneously we get
$\text{A}=\frac{1}{5},\text{ B}=-\frac{1}{5},\text{ C}=\frac{2}{5}$
$\text{I}=\int\bigg(\frac{1}{5\text{x}+1}+\frac{\frac{-1}{5}\text{x}+\frac{2}{5}}{\text{x}^2+1}\bigg)\text{dx}$
Integrating we get as,
$\frac{1}{5}\log|\text{x}+2|-\frac{1}{10}\log|\text{x}^2+1|+\frac{2}{5}\tan^{-1}\text{x}+\text{C}$
$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$

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Question 1681 Mark

$\int\frac{1}{7}\sin\Big(\frac{\text{x}}{7}+10\Big)\text{dx}$ is equal to:

$\frac{1}{7}\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$
$-\frac{1}{7}\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$
$\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$
$-7\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$
$\cos(\text{x}+70)+\text{C}$

Answer

$\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$

Solution:
Let $\text{I}=\int\frac{1}{7}\sin\Big(\frac{\text{x}}{7}+10\Big)\text{dx}$
$=\frac{1}{7}\int\sin\Big(\frac{\text{x}}{7}+10\Big)=\frac{1}{7}\frac{-\cos\Big(\frac{\text{x}}{7}+10\Big)}{\frac{1}{7}}$
$=-\cos\Big(\frac{\text{x}}{7}+10\Big)+\text{C}$

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Question 1691 Mark

Choose the correct answer in Exercise: The value of $\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\bigg)\text{dx}\ $is

2
$\frac{3}{4}$
0
-2

Answer

$\text{Let}\ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\bigg)\text{dx}\ \ \ ....(1)$
$\Rightarrow\ \ \ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\log\begin{bmatrix} \frac{4+3\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)}{4+3\cos\bigg(\frac{\pi}{2}-\text{x}\bigg)}\text{dx}\\ \end{bmatrix}\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg[\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\bigg]\text{dx}\ \ \ ....(2)$
Adding eq. (i) and (ii),
$2\text{I}=\int^{\frac{\pi}{2}}\limits_{0}\bigg[\log\bigg(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\bigg)+\log\bigg(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\bigg)\bigg]\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\bigg[\log\bigg(\frac{4+3\sin\text{x}}{4+3\sin\text{x}}\bigg)\bigg(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\bigg)\bigg]\text{dx}$
$\Rightarrow\ \ \ 2\text{I}=\int^{\frac{\pi}{2}}\limits_{0}(\log1)\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}0\ \text{dx}=0\ \ \ \ \ \Rightarrow\text{I}=0$

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Question 1701 Mark

Choose the correct answer in Exercise:
If $\text{f (a}+\text{b)}-\text{x}=\text{f (x)},$ then $\int^{\text{b}}_{\text{a}}\text{x f (x)}\ \text{dx}$ is equal to

$\frac{\text{a}+\text{b}}{2}\int^{\text{a}}_{\text{b}}\text{f (b}-\text{x)}\text{dx}$
$\frac{\text{a}+\text{b}}{2}\int^{\text{b}}_{\text{a}}\text{f (b}+\text{x)}\text{dx}$
$\frac{\text{b}-\text{a}}{2}\int^{\text{b}}_{\text{a}}\text{f (x)}\text{dx}$
$\frac{\text{a}+\text{b}}{2}\int^{\text{b}}_{\text{a}}\text{f (x)}\text{dx}$

Answer

$\frac{\text{a}+\text{d}}{2}\int^{\text{b}}\limits_{\text{a}}\text{f (x)}\text{dx}$

$\text{Let I}=\int^{\text{b}}\limits_{a}\text{x f (x)}\text{dx}$
$\text{I}=\int^{\text{b}}\limits_{\text{a}}\text{(a}+\text{b}-\text{x)}\text{f (a}+\text{b}-\text{x)}\text{dx}\ \ \ \ \ \Bigg[\int^{\text{b}}\limits_{\text{a}}\text{f (x)dx}=\int^{\text{b}}\limits_{\text{a}}\text{f (a}+\text{b}-\text{x)}\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int^{\text{b}}\limits_{\text{a}}\text{(a}+\text{b}-\text{x)}\ \text{f (x)}\text{dx}$
$\Rightarrow\text{I}=\text{(a}+\text{b)}\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}\ \ \ -\text{I}$ [Using(1)]
$\Rightarrow\text{I}+\text{I}=\text{(a}+\text{b)}\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}$
$\Rightarrow2\text{I}=\text{(a}+\text{b)}\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}$
$\Rightarrow\text{I}=\bigg(\frac{\text{a}+\text{b}}{2}\bigg)\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}$

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Question 1711 Mark

$\int\frac{\cos2\text{x}-1}{\cos2\text{x}+1}\text{ dx}=$

$\tan\text{x}-\text{x}+\text{C}$
$\text{x}+\tan\text{x}+\text{C}$
$\text{x}-\tan\text{x}+\text{C}$
$-\text{x}-\cot\text{x}+\text{C}$

Answer

$\text{x}-\tan\text{x}+\text{C}$

Solution:
$\text{I}=\int\frac{\cos2\text{x}-1}{\cos2\text{x}+1}\text{ dx}$
$\text{I}=-\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{ dx}$
$\text{I}=-\int\frac{2\sin^2\text{x}}{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$\text{I}=-\int\tan^2\text{x dx}$
$\text{I}=-\int(\sec^2\text{x}-1)\text{dx}$
$\text{I}=-(\tan\text{x}-\text{x})+\text{C}$
$\text{I}=\text{x}-\tan\text{x}+\text{C}$

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Question 1721 Mark

The value of $\int\frac{\text{d}(\text{x}^2+1)}{\sqrt{\text{x}^2+2}}$ is:

$2\sqrt{\text{x}^2+2}+\text{c}$
$\sqrt{\text{x}^2+2}+\text{c}$
$\text{x}\sqrt{\text{x}^2+2}+\text{c}$
${4}\sqrt{\text{x}^2+2}+\text{c}$

Answer

$2\sqrt{\text{x}^2+2}+\text{c}$

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Question 1731 Mark

The value of $\int\limits^\frac{\pi}{4}_0\cos\text{x}\text{ e}^{\sin\text{x}}\text{ dx}$ is:

1
e - 1
0
1

Answer

$\text{e}-1$

Solution:
Let, $\text{I}=\int\limits^\frac{\pi}{4}_0\cos\text{x}\text{ e}^{\sin\text{x}}\text{dx}$
Let $\sin\text{x}=\text{t},$ then $\cos\text{x}\text{ dx}=\text{dt}$
when $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore the integrel becomes
$\text{I}=\int\limits^1_0\text{e}^\text{t}\text{dt}$
$=\big[\text{e}^\text{t}\big]^1_0$
$=\text{e}-1$

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Question 1741 Mark

$\int\big(1+5\text{x}+10\text{x}^2+10\text{x}^3+5\text{x}^4+\text{x}^5\big)\text{dx}=\frac{(1+\text{x})\text{p}}{6}+\text{c}$ then p is:

Answer

Solution:
Given, $\int\big(1+5\text{x}+10\text{x}^2+10\text{x}^3+5\text{x}^4+\text{x}^5\big)\text{dx}$
$\int(\text{x+1})^5\text{dx}$ [by binomial theorem]
$=(\text{x+1})^5\text{dx}=\frac{(\text{x+1})6}{6}+\text{c}$
So, the value of p = 6

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Question 1751 Mark

$\int\text{e}^{\text{x}}(1-\cot\text{x}+\cot^2\text{x})\text{dx}=$

$\text{e}^{\text{x}}\cot\text{x}+\text{C}$
$-\text{e}^{\text{x}}\cot\text{x}+\text{C}$
$\text{e}^{\text{x}}\text{cosec x}+\text{C}$
$-\text{e}^{\text{x}}\text{cosec x}+\text{C}$

Answer

$-\text{e}^{\text{x}}\cot\text{x}+\text{C}$

Solution:
$\text{I}=\int\text{e}^{\text{x}}(1-\cot\text{x}+\cot^2\text{x})\text{dx}$
$\text{I}=\int\text{e}^\text{x}(1+\cot^2\text{x}-\cot\text{x})\text{dx}$
$\text{I}=\int\text{e}^{\text{x}}(\text{cosec}^2\text{x}-\cot\text{x})\text{dx}$
Here, $\text{f(x)}=-\cot\text{x}$
$\text{f}'(\text{x})=\text{cosec}^2\text{x}$
$\text{I}=-\text{e}^{\text{x}}\cot\text{x}+\text{C}$

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Question 1761 Mark

Evaluate: $\int\frac{1}{1+3\sin^2\text{x}+8\cos^2\text{x}}\text{dx}.$

$\frac{1}{6}\tan^{-1}(2\tan\text{x})+\text{c}$
$\tan^{-1}(2\tan\text{x})+\text{c}$
$\frac{1}{6}\tan^{-1}(\frac{2\tan\text{x}}{3})+\text{c}$
$\text{None of these}$

Answer

$\frac{1}{6}\tan^{-1}(\frac{2\tan\text{x}}{3})+\text{c}$

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Question 1771 Mark

$\int\frac{\text{xdx}}{(\text{x-1)}(\text{x-2})}$ equals:

$\log\begin{vmatrix}\frac{(\text{x}-1)^2}{\text{x}-2}\end{vmatrix}+\text{c}$
$\log\begin{vmatrix}\frac{(\text{x}-2)^2}{\text{x}-2}\end{vmatrix}+\text{c}$
$\log\begin{vmatrix}\Big(\frac{\text{x}-1}{\text{x}-2}\Big)^2\end{vmatrix}+\text{c}$
$\log|(\text{x}-1)(\text{x}-2)+\text{c}$

Answer

$\log\begin{vmatrix}\frac{(\text{x}-2)^2}{\text{x}-2}\end{vmatrix}+\text{c}$

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Question 1781 Mark

$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$ equals:

$0$
$\frac{1}{2}$
$2$
$\frac{3}{2}$

Answer

Solution:
$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$
$=\int\limits^\pi_0\frac{1}{1+\sin\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}\text{dx}$
$= \int\limits^\pi_0\frac{1-\sin\text{x}}{1-\sin^2\text{x}}\text{dx}$
$= \int\limits^\pi_0\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{dx}$
$=\int\limits^\pi_0(\sec^2\text{x}-\sec\text{x}\tan\text{x})\text{dx}$
$=\big[\tan\text{x}-\sec\text{x}\big]^\pi_0$
$=0+1-0+1$
$=2$

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Question 1791 Mark

The primitive of the function $\text{f(x)}=\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}},\text{ a}>0$ is:

$\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$
$\log_\text{e}\text{a}\cdot\text{a}^{\text{x}+\frac{1}{\text{x}}}$
$\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\text{x}}{\log_\text{e}\text{a}}$
$\text{x}\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$

Answer

$\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$

Solution:
$\text{f(x)}=\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}}$
$\therefore\ \int\text{f(x)}\text{dx}=\int\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}}$
Let $\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{t}$
$\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\ \int\text{f(x)}\text{dx}=\int\text{a}^{\text{t}}\text{ dt}$
$=\frac{\text{a}^{\text{t}}}{\log_\text{e}\text{a}}+\text{C}$
$=\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}+\text{C}$ $\Big(\because\text{t}=\text{x}+\frac{1}{\text{x}}\Big)$

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Question 1801 Mark

If $\frac{\text{dy}}{\text{dx}}=3$ then y is equal to:

3x
0
3x + c
$\frac{\text{x}}{3}+\text{c}$

Answer

3x + c

Solution:
$\frac{\text{dy}}{\text{dx}}=3$
$\text{dy}=3\text{dx}$
$\int\text{dy}=\int3\text{dx}=3\text{x}+\text{c}$
$\therefore\text{y}=3\text{x}+\text{c}$

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Question 1811 Mark

The value of $\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)\text{dx}$ is:

2
$\frac{3}{4}$
0
-2

Answer

Solution
Let $\text{I}=\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)\text{dx}\ ...(\text{i})$
$=\int\limits^\frac{\pi}{2}_0\log\Bigg[\frac{4+3\sin\big(\frac{\pi}{2}-\text{x}\big)}{4+3\cos\big(\frac{\pi}{2}-\text{x}\big)}\Bigg]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\Big)\text{dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^\frac{\pi}{2}_0\Big[\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)+\log\Big(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\Big)\Big]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\log1\text{ dx}=0$
Hence, $\text{I}=0$

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Question 1821 Mark

Choose the correct option from given four options:
$\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}$ is equal to:

$\frac{1}{5\text{x}}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
$\frac{1}{5}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
$\frac{1}{10\text{x}}(1+4)^{-5}+\text{C}$
$\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$

Answer

$\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$

Solution:
Let $\text{I}=\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}=\int\frac{\text{x}^9}{\text{x}^{12}\Big(4+\frac{1}{\text{x}^2}\Big)^6}\text{dx}=\int\frac{\text{dx}}{\text{x}^3\Big(4+\frac{1}{\text{x}^2}\Big)^6}$
Put $4+\frac{1}{\text{x}^2}=\text{t}\Rightarrow\frac{-2}{\text{x}^3}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^6}=-\frac{1}{2}\Big[\frac{\text{t}^{-6+1}}{-6+1}\Big]+\text{C}$
$=\frac{1}{10}\Big[\frac{1}{\text{t}^5}\Big]+\text{C}=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$

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Question 1831 Mark

The value of $\int\limits^1_0\tan^{-1}\Big(\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\Big)\text{ dx},$ is:

1
0
-1
$\frac{\pi}{4}$

Answer

Solution:
Let, $\text{I}=\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{i})$
$=\int\limits^1_0\tan^{-1}\frac{2(1-\text{x})-1}{1+(1-\text{x})-(1-\text{x})^2}\text{ dx}$
$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{2-\text{x}-1-\text{x}^2+2\text{x}}\text{ dx}$
$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{1+\text{x}-\text{x}^2}\text{ dx}$
$=-\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}-\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}$
Hence, $\text{I}=0$

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Question 1841 Mark

$\int\frac{2\text{x}\log(1+\text{x}^2)}{1+\text{x}^2}\text{dx:}$

$\log(1+\text{x}^2)+\text{c}$
$\frac{[\log(1+\text{x}^2)^2]}{2}+\text{c}$
$2\log(1+\text{x}^2)+\text{c}$
None of these

Answer

$\frac{[\log(1+\text{x}^2)^2]}{2}+\text{c}$

Solution:
$\int\frac{2\text{x}\log(1+\text{x}^2)}{1+\text{x}^2}\text{dx}$
Let $\log(1+\text{x}^2) =\text{z}$ and $\frac{\text{2x}}{1+\text{x}^2}\text{dx}=\text{dz}$ Using these in the above integration we get,
$=\int\text{z, }{\text{dz}}$
$\frac{\text{z}^2}{2}+\text{c}$ [Where c is integrating constant]
$=\frac{[\log(1+\text{x}^2)^2]}{2}+\text{c}$

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MCQ 1851 Mark

$\int\limits^3_0\frac{3\text{x}+1}{\text{x}^2+9}\text{ dx}=$

✓
$\frac{\pi}{12}+\log\big(2\sqrt{2}\big)$
B
$\frac{\pi}{2}+\log\big(2\sqrt{2}\big)$
C
$\frac{\pi}{6}+\log\big(2\sqrt{2}\big)$
D
$\frac{\pi}{3}+\log\big(2\sqrt{2}\big)$

Answer

Correct option: A.

$\frac{\pi}{12}+\log\big(2\sqrt{2}\big)$

We have,
$\text{I}=\int\limits^3_0\frac{3\text{x}+1}{\text{x}^2+9}\text{ dx}$
$\text{I}=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$\text{I}_1=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}$ and $\text{I}_2=\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
Putting $\text{x}^2+9=\text{t}$ in $I_1$
$\Rightarrow 2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{x}\text{ dx}=\frac{\text{dt}}{2}$
when $\text{x}\rightarrow0;\text{t}\rightarrow9$
and $\text{x}\rightarrow3;\text{t}\rightarrow18$
$\therefore\text{I}=\int\limits^{18}_9\frac{3\text{ dt}}{2\text{ t}}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$=\frac{3}{2}\int\limits^{18}_9\frac{\text{dt}}{\text{t}}+\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$
$=\frac{3}{2}\big[\log(\text{t})\big]^{18}_9+\frac{1}{3}\Big[\tan^{-1}\Big(\frac{\pi}{3}\Big)\Big]^3_0$
$=\frac{3}{2}\big[\log18-\log9\big]+\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$
$=\frac{3}{2}\Big[\log\frac{18}{9}\Big]+\frac{\pi}{12}$
$=\frac{3}{2}\big[\log2\big]+\frac{\pi}{12}$
$=\log(\sqrt{8})+\frac{\pi}{12}$
$=\log(2\sqrt{2})+\frac{\pi}{12}$
$=\frac{\pi}{12}+\log(2\sqrt{2})$

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Question 1861 Mark

What is the value of $\int_{0}^{\frac{\pi}{2}}\frac{\sin\text{x}-\cos\text{x}}{1+\sin\text{x}\cos\text{x}}\text{dx}?$

$1$
$\frac{\pi}{2}$
$0$
$-\frac{\pi}{2}$

Answer

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Question 1871 Mark

$\int\frac{\text{x+sin x}}{1+\text{cos x}}$ dx is equal to:

$\log|1+\cos\text{x}|+\text{c}$
$\log|\text{x}+\sin\text{x}|+\text{c}$
$\text{x}-\tan+\text{c}$
$\text{x}.\text{tan}\frac{\text{x}}{2}+\text{c}$

Answer

$\text{x}.\text{tan}\frac{\text{x}}{2}+\text{c}$

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Question 1881 Mark

$\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sin2\text{x}}\text{ dx}$ is equal to:

$\log_\text{e}{3}$
$\log_\text{e}\sqrt{3}$
$\frac{1}{2}\log(-1)$
$\log(-1)$

Answer

$\log_\text{e}\sqrt{3}$

Solution:
$\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sin2\text{x}}\text{ dx}$
$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\text{cosec }2\text{x}\text{ dx}$
$=\frac{1}{2}\int\limits^\frac{\pi}{3}_\frac{\pi}{6}2\text{cosec }2\text{x}\text{ dx}$
$=\frac{-1}{2}\big[\log(\text{cosec}2\text{x}+\cot2\text{x})\big]^\frac{\pi}{3}_\frac{\pi}{6}$
$=\frac{-1}{2}\big[-2\log\sqrt{3}\big]$
$=\log\sqrt{3}$

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Question 1891 Mark

$\int\frac{\cot\text{x}}{3\sqrt{\sin\text{x}}}\text{dx}=$

$\frac{-3}{3\sqrt{sin\text{x}}}+\text{c}$
$\frac{-2}{\sqrt{sin^3\text{x}}}+\text{c}$
$\frac{3}{\sqrt{sin^{\frac{1}{3}}\text{x}}}+\text{c}$
$\text{None of there}$

Answer

$\frac{-3}{3\sqrt{sin\text{x}}}+\text{c}$

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Question 1901 Mark

$\int\limits^\pi_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}=$

$\sqrt{1-\pi^2}-1$
$\frac{\pi}{2}-1$
$\frac{\pi}{2}+1$
${\pi}+{1}$

Answer

$\sqrt{1-\pi^2}-1$

Solution:
We have,
$\text{I}=\int\limits^\pi_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$=\int\limits^\pi_0\Big[\sqrt{\frac{1-\text{x}}{1+\text{x}}}\times\frac{\sqrt{1-\text{x}}}{\sqrt{1-\text{x}}}\Big]\text{dx}$
$=\int\limits^\pi_0\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
$=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}-\int\limits^\pi_0\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Putting $1 -\text{x}^2=\text{t}$
$\Rightarrow - 2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{x}\text{ dx}=\frac{-\text{dt}}{2}$
when $\text{x}\rightarrow0;\text{t}\rightarrow1$
and $\text{x}\rightarrow\pi;\text{ t}\rightarrow-\pi^2$
$\therefore\ \text{I}=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}-\int\limits^{(1-\pi^2)}_1\frac{-\text{dt}}{2\sqrt{\text{t}}}$
$=\big[\sin^{-1}\text{x}\big]^{\pi}_0+\frac{2}{2}\big[\sqrt{\text{t}}\big]^{1-\pi^2}_1$
$=\big[0-0\big]+\Big[\sqrt{1-\pi^2}-\sqrt{1}\Big]$
$=\sqrt{1-\pi^2}-1$

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Question 1911 Mark

$\int_{0}^{1}\frac{(\tan^{-1}\text{x})^2}{1+\text{x}^2}\text{dx}=$

$1$
$\frac{\pi^2}{64}$
$\frac{\pi^2}{192}$
$\text{None of these}$

Answer

$\frac{\pi^2}{192}$

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Question 1921 Mark

Choose the correct option from given four options:
$\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\frac{\text{dx}}{1+\cos2\text{x}}$ is equal to:

Answer

Solution:
We have, $\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\frac{\text{dx}}{1+\cos2\text{x}}=\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\frac{\text{dx}}{2\cos^2\text{x}}$ $[\because1+\cos2\text{A}=2\cos^2\text{A}]$
$=\frac{1}{2}\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\sec^2\text{x dx}$
$=\int\limits^{\frac{\pi}{4}}_0\sec^2\text{x dx}$ $\int\limits^\text{a}_{-\text{a}}\text{f(x)dx}=\begin{cases}\int\limits^\text{a}_0\text{f(x)},&\text{if f(x) is even}\\0,&\text{if f(x) is odd}\end{cases}$
$=\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_0=1$
$=\tan\frac{\pi}{4}-\tan0$
$=1$

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Question 1931 Mark

$\int\frac{\text{dx}}{\sin(\text{x-a})\sin(\text{x-b})}$ is equal to:

$\sin(\text{b-a})\log|\frac{\sin(\text{x-b})}{\sin(\text{x-a})}|+\text{c}$
$\text{cosec}(\text{b-a})\log|\frac{\sin(\text{x-b})}{\sin(\text{x-b})}|+\text{c}$
$\text{cosec}(\text{b-a})\log|\frac{\sin(\text{x-b})}{\sin(\text{x-a})}|+\text{c}$
$\sin(\text{b-a})\log|\frac{\sin(\text{x-a})}{\sin(\text{x-b})}|+\text{c}$

Answer

$\text{cosec}(\text{b-a})\log|\frac{\sin(\text{x-b})}{\sin(\text{x-a})}|+\text{c}$

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Question 1941 Mark

Solve $\int\frac{1}{\sqrt{9-25\text{x}^2}}\text{dx}$

$\frac{1}{5}\sin^{-1}\big(\frac{5\text{x}}{3}\big)+\text{c}$
$\sin^{-1}\big(\frac{5\text{x}}{3}\big)+\text{c}$
$\frac{1}{5}\sin^{-1}\big(\frac{3\text{x}}{3}\big)+\text{c}$
$\sin^{-1}\big(\frac{3\text{x}}{3}\big)+\text{c}$

Answer

$\frac{1}{5}\sin^{-1}\big(\frac{5\text{x}}{3}\big)+\text{c}$

Solution:
We have,
$\text{I}=\int\frac{\text{dx}}{\sqrt{9-25\text{x}^2}}$
$\text{I}=\int\frac{\text{dx}}{5\sqrt\frac{{9}}{{25}}\text{-x}^2}$
$\text{I}=\frac{1}{5}\int\frac{\text{dx}}{\sqrt{\frac{{3}}{{5}}^2\text{-x}^2}}$
We know that
$\int\frac{\text{dx}}{\text{a}^2-\text{x}^2}=\sin^{-1}\big(\frac{\text{x}}{\text{a}}\big)+\text{C}$
$\therefore\text{I} = \frac{1}{5}\sin^{-1}\Big(\frac{\text{x}}{\frac{{3}}{5}}\Big)+\text{C}$
$\therefore\text{I} = \frac{1}{5}\sin^{-1}\Big(\frac{5\text{x}}{3}\Big)+\text{C}$

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Question 1951 Mark

If $\text{f}'(\text{x})=\text{x}+\frac{1}{\text{x}},$ then value of f(x) is:

$\text{x}^2+\log\text{x + c}$
$\frac{\text{x}^2}{2}+\log\text{x + c}$
$\frac{\text{x}}{2}+\log\text{x + c}$
None of these

Answer

$\frac{\text{x}^2}{2}+\log\text{x + c}$

Solution:
Given,
$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$
On integrating both sides, we get
$\text{f}(\text{x})=\frac{\text{x}^{2}}{{2}}+\log\text{x + c}$

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Question 1961 Mark

$\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}$ is equal to:

$\frac{1}{5\text{x}}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
$\frac{1}{5}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
$\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$

Answer

$\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$

Solution:
Let $\text{I}=\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}$
$=\int\frac{\text{x}^{9}}{\text{x}^{\frac{1}{2}}\big(4+\frac{1}{\text{x}^2}\big)^6}\text{ dx}$
$=\int\frac{\text{x}^{\frac{1}{3}}}{\big(4+\frac{1}{\text{x}^2}\big)^6}\text{ dx}$
Let $\Big(4+\frac{1}{\text{x}^2}\Big)=\text{t}$
On differentiating both sides, we get
$-\frac{2}{\text{x}^3}\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\frac{1}{2}\int\frac{1}{(\text{t}^{6})}\text{dt}$
$=-\frac{1}{2}\Big(-\frac{1}{5}\Big)\text{t}^{-5}+\text{C}$
$=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
Therefore, $\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
Hence, the correct option is (d)

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Question 1971 Mark

$\int\sin^{-1}\text{xdx}$ is equal to:

$\cos^{-1}\text{x}+\text{c}$
$\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}+\text{c}$
$\frac{1}{\sqrt{1-\text{x}^2}}+\text{c}$
$\text{x}\sin^{-1}\text{x}-\sqrt{1-\text{x}^2}+\text{c}$

Answer

$\text{x}\sin^{-1}\text{x}-\sqrt{1-\text{x}^2}+\text{c}$

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Question 1981 Mark

The number of integral solutions (x, y) of the equations $\text{x}{\sqrt{\text{y}}}+\text{y}\sqrt{\text{x}}=20$ and $\text{x}{\sqrt{\text{x}}}+\text{y}\sqrt{\text{y}}=65$ is:

0
1
2
None of these

Answer

Solution:
By trial and error method Put y = 1, x = 16 in the same way x = 1, y = 16 The equation gets satisfied.

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Question 1991 Mark

$\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$ equals:

$\frac{1}{3}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
$\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
${\sqrt{3 }}\tan^{-1}\big({\sqrt{3}}\big)$
$2{\sqrt{3 }}\tan^{-1}{\sqrt{3}}$

Answer

$\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$

Solution:
We have,
$=\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{1}{2+\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{1+\tan^2\frac{\text{x}}{2}}{2+2\tan^2\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{\sec^2\frac{\text{x}}{2}}{3+\tan^2\frac{\text{x}}{2}}\text{dx}$
Putting $\tan\frac{\text{x}}{2}=\text{t}$
$\Rightarrow \frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$\Rightarrow \sec^2\frac{\text{x}}{2}\text{dx}=2\text{dt}$
when $\text{x}\rightarrow0;\text{ t}\rightarrow0$
and $\text{x}\rightarrow\frac{\pi}{2};\text{ t}\rightarrow1$
$\therefore\ \text{I}=\int\limits^1_0\frac{2}{3+\text{t}^2}\text{ dt}$
$=2\int\limits^1_0\frac{1}{(\sqrt{3})^2+\text{t}^2}\text{dt}$
$=\frac{2}{\sqrt{3}}\Big[\tan^{-1}\frac{\text{t}}{\sqrt{3}}\Big]^1_0$
$=\frac{2}{\sqrt{3}}\Big[\tan^{-1}\frac{1}{\sqrt{3}}-\tan^{-1}\frac{0}{\sqrt{3}}\Big]$
$=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$

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Question 2001 Mark

Given that $\int\limits^{\infty}_0\frac{\text{x}^2}{(\text{x}^2+\text{a}^2)(\text{x}^2+\text{b}^2)(\text{x}^2+\text{c}^2)}\text{ dx}=\frac{\pi}{2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})},$ the value of $\int\limits^\infty_0\frac{1}{(\text{x}^2+4)(\text{x}^2+9)},$ is:

$\frac{\pi}{60}$
$\frac{\pi}{20}$
$\frac{\pi}{40}$
$\frac{\pi}{80}$

Answer

$\frac{\pi}{60}$

Solution:
$\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)\big(\text{x}^2+9\big)}\text{dx}$
$=\frac{1}{5}\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)}-\frac{1}{\big(\text{x}^2+9\big)}\text{dx}$
$=\frac{1}{5}\bigg[\frac{1}{2\tan^{-1}{}}\frac{\text{x}}{2}-\frac{1}{3}\tan^{-1}\frac{\text{x}}{3}\bigg]^\infty_0$
$=\frac{1}{5}\bigg[\frac{1}{2}\times\frac{\pi}{2}-\frac{1}{3}\times\frac{\pi}{2}\bigg]$
$=\frac{1}{5}\times\frac{\pi}{12}$
$=\frac{\pi}{60}$

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MATHS M.C.Q (1 Marks)