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MATHS M.C.Q (1 Marks)

INTEGRALS · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 210 questions.

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M.C.Q (1 Marks)

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Question 11 Mark
$\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2\text{x dx}$ is equal to:
  1. -1
  2. 0
  3. 1
  4. 2
Answer
  1. 2
Solution:
$\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2\text{x dx}$
$\Rightarrow\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2\text{x dx}$ $=\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}$
$\Rightarrow\Big[\tan\big(\frac{\pi}{4}\big)-\tan\big(-\frac{\pi}{4}\big)\Big]$
$\Rightarrow[1-(-1)]$
$\Rightarrow2$
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Question 21 Mark
Choose the correct answer in Exercise:
The value of the integral   $\int^{1}_{\frac{1}{3}}\frac{(\text{x}-\text{x}^{3})^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}\ \text{is}$
  1. 6
  2. 0
  3. 3
  4. 4
Answer
  1. 6
$\text{Let}\ \text{I}=\int\limits_{\frac{1}{3}}^{1}\frac{(\text{x}-\text{x}^{3})^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}=\int\limits_{3}^{1}\frac{\text{x}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{4}}\text{dx} $
$=\int\limits_{\frac{1}{3}}^{1}\frac{\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{3}}\text{dx}=\int\limits_{\frac{1}{3}}^{1}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}.\frac{1}{\text{x}^{3}}\text{dx}$
$\text{put}\frac{1}{\text{x}^{2}}=\text{t},\ \therefore-\frac{2}{\text{x}^{3}}\text{dx}=\text{dt}\Rightarrow\frac{1}{\text{x}^{3}}\text{dx}=-\frac{1}{2}\text{dt}$
$\text{when}\ \text{x}=\frac{1}{3},\text{t}=9$
$\text{when}\ \text{x}=1,\text{t}=1$
$\therefore\ \ \text{I}=-\frac{1}{2}\int^{1}_{9}\big(\text{t}-1\big)^{\frac{1}{3}}\text{dt}=-\frac{1}{2}\Bigg[\frac{\big(\text{t}-1\big)^{\frac{4}{3}}}{\frac{4}{3}}\Bigg]^{1}_{0}$
$=-\frac{3}{8}\bigg[\big(\text{t}-1\big)^{\frac{4}{3}}\bigg]^{1}_{9}=-\frac{3}{8}\bigg[\big(1-1\big)^{\frac{4}{3}}-\big(9-1\big)^{\frac{4}{3}}\bigg]$
$=-\frac{3}{8}\bigg[{0}-(8)^{\frac{4}{3}}\bigg]^{1}_{9}=\frac{3}{8}(2)^{3^{\frac{4}{3}}}=\frac{3}{8}\text{x}\ 2^{4}=\frac{3}{8}\text{x}16=6$
$\text{Let}\ \text{I}=\int^{1}_{\frac{1}{3}}\frac{\big(\text{x}-\text{x}^{3}\big)^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}=\int^{1}_{3}\frac{\text{x}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}$
$=\int^{1}_{\frac{1}{3}}\frac{\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{3}}\text{dx}=\int^{1}_{\frac{1}{3}}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}.\frac{1}{\text{x}^{3}}\text{dx}$
$\text{put}\frac{1}{\text{x}^{2}}=\text{t},\ \therefore-\frac{2}{\text{x}^{3}}\text{dx}=\text{dt}\Rightarrow\frac{1}{\text{x}^{3}}\text{dx}=-\frac{1}{2}\text{dt}$
$\text{when}\ \text{x}=\frac{1}{3},\text{t}=9$
$\text{when}\ \text{x}=1,\text{t}=1$
$\therefore\ \ \text{I}=-\frac{1}{2}\int^{1}_{9}\big(\text{t}-1\big)^{\frac{1}{3}}\text{dt}=-\frac{1}{2}\bigg[\frac{\text{(t}-1)^{\frac{4}{3}}}{\frac{4}{3}}\bigg]^{1}_{0}$
$=-\frac{3}{8}\bigg[(\text{t}-1)^{\frac{4}{3}}\bigg]^{1}_{9}=-\frac{3}{8}\bigg[(1-1)^{\frac{4}{3}}-(9-1)^{\frac{4}{3}}\bigg]$
$=-\frac{3}{8}\bigg[0-(8)^{\frac{4}{3}}\bigg]=-\frac{3}{8}(2)^{3^{\frac{4}{3}}}=\frac{3}{8}\times2^{4}=\frac{3}{8}\times16=6$
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Question 31 Mark
$\int\frac{(1+\text{log x})^2}{1+\text{x}^2}\text{dx}=$
  1. $\frac{1}{3}(1+\text{log})^3+\text{c}$
  2. $\frac{1}{2}(1+\text{log})^2+\text{c}$
  3. $\log(\text{log }1+\text{x})+2$
  4. $\text{None of these}$
Answer
  1. $\frac{1}{3}(1+\text{log})^3+\text{c}$
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Question 41 Mark
The value of the integral $\int\limits^2_{-2}\big|1-\text{x}^2\big|\text{dx}$ is:
  1. 4
  2. 2
  3. -2
  4. 0
Answer
  1. 4
Solution:
We have,
$\text{I}=\int\limits^2_{-2}\big|1-\text{x}^2\big|\text{dx}$
$\big|1-\text{x}^2\big|=\begin{cases}-\big(1-\text{x}^{2}\big), & -2<\text{x}<-1\\\big(1-\text{x}^2\big), & -1<\text{x}<1 \\-\big(1-\text{x}^2\big),&1<\text{x}<2\end{cases}$
$\therefore\ \text{I}=\int\limits^{-1}_{-2}\big|1-\text{x}^2\big|\text{dx}+\int\limits^1_{-1}\big|1-\text{x}^2\big|\text{dx}+\int\limits^2_1\big|1-\text{x}^2\big|\text{dx}$
$=\int\limits^{-1}_{-2}-\big(1-\text{x}^2)\text{dx}+\int\limits^1_{-1}(1-\text{x}^2)\text{dx}+\int\limits^2_1-(1-\text{x}^2)\text{dx}$
$=-\int\limits^{-1}_{-2}(1-\text{x}^2)\text{dx}+\int\limits^1_{-1}(1-\text{x}^2)\text{dx}-\int\limits^2_1(1-\text{x}^2)\text{dx}$
$=\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^{-1}_{-2}+\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^1_{-1}-\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^2_1$
$=\Big[-1+\frac{1}{3}+2-\frac{8}{3}\Big]+\Big[1-\frac{1}{3}+1-\frac{1}{3}\Big]-\Big[2-\frac{8}{3}-1+\frac{1}{3}\Big]$
$=-\Big[1-\frac{7}{3}\Big]+\Big[2-\frac{2}{3}\Big]-\Big[1-\frac{7}{3}\Big]$
$=-1+\frac{7}{3}+2-\frac{2}{3}-1+\frac{7}{3}$
$=4$
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Question 51 Mark
$\int\frac{\text{x}^4+1}{\text{x}^2+1}\text{ dx}$  is equal to:
  1. $\frac{\text{x}^3}{3}+\text{x}+\tan^{-1}\text{x}+\text{c}$
  2. $\frac{\text{x}^3}{3}-\text{x}+\tan\text{x}+\text{c}$
  3. $\frac{\text{x}^3}{3}+\text{x}+2\tan^{-1}\text{x}+\text{c}$
  4. $\frac{\text{x}^3}{3}-\text{x}+2\tan^{-1}\text{x}+\text{c}$
Answer
  1. $\frac{\text{x}^3}{3}-\text{x}+2\tan^{-1}\text{x}+\text{c}$
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Question 61 Mark
$\int\frac{2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\text{ dx}=$
  1. $\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
  2. $-\frac{1}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
  3. $\frac{-1}{(\text{e}^{\text{x}}+1)^2}+\text{C}$
  4. $\frac{1}{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}+\text{C}$
Answer
  1. $\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
Solution:
$\text{I}=\int\frac{2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\text{ dx}$
$\text{I}=\int\frac{2\text{e}^{2\text{x}}}{(\text{e}^{2\text{x}}+1)^2}\text{ dx}$
Put $\text{t}=\text{e}^{2\text{x}}+1$
$\text{dt}=2\text{e}^{2\text{x}}\text{ dx}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2}$
$\text{I}=\frac{-1}{\text{t}}+\text{C}$
$\text{I}=\frac{-1}{\text{e}^{2\text{x}}+1}+\text{C}$
$\text{I}=\frac{-\frac{1}{\text{e}^{\text{x}}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
$\text{I}=\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
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Question 71 Mark
$\int_{0}^{\infty}\frac{1}{1+\text{e}^\text{x}}\text{dx}=$
  1. $\log2$
  2. $-\log2$
  3. $\log2-1$
  4. $\log4-1$
Answer
  1. $\log2$
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Question 81 Mark
The value of the integral $\int\limits^\infty_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{dx}$ is:
  1. $\frac{\pi}{2}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{6}$
  4. $\frac{\pi}{3}$
Answer
  1. $\frac{\pi}{4}$
Solution:
We have, 
$\text{I}=\int\limits^\infty_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{ dx}$
Putting $\text{x}=\tan\theta$
$\Rightarrow \text{dx}=\sec^2\theta\text{ d}\theta$
when $\text{x}\rightarrow0;\theta\rightarrow0$
and $\text{x}\rightarrow\infty;\theta\rightarrow\frac{\pi}{2}$
Now, integral becomes
$\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\tan\theta}{(1+\tan\theta)\sec^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\tan\theta}{1+\tan\theta}\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\frac{\sin\theta}{\cos\theta}}{1+\frac{\sin\theta}{\cos\theta}}\text{d}\theta$
$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\theta}{\sin\theta+\cos\theta}\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\big(\frac{\pi}{2}-\theta\big)}{\sin\big(\frac{\pi}{2}-\theta\big)+\cos\big(\frac{\pi}{2}-\theta\big)}\text{ d}\theta$ $\bigg[\therefore\ \int\limits^\text{a}_0\text{f}(\text{x)}\text{dx}=\int\limits^\text{a}_0\text{f}(\text{a}-\text{x})\text{dx}\bigg]$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\theta}{\cos\theta+\sin\theta}\text{d}\theta$
$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\theta}{\sin\theta+\cos\theta}\text{d}\theta\ ....(\text{ii})$
Adding (i) and (ii), we get
$2\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\theta+\cos\theta}{\sin\theta+\cos\theta}\text{d}\theta$
$\Rightarrow 2\text{I}=\int\limits^\frac{\pi}{2}_0\text{d}\theta$
$\Rightarrow 2\text{I}=\frac{\pi}{2}$
$\Rightarrow \text{I}=\frac{\pi}{4}$
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Question 91 Mark
If $\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$ then y =
  1. $\text{ln }\text{x}+\text{c}$
  2. $\text{x}+\text{c}$
  3. $\frac{-1}{\text{x}^2}+\text{c}$
  4. $\frac{1}{\text{x}^2}+\text{c}$
Answer
  1. $\text{ln }\text{x}+\text{c}$
Solution:
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\text{dy}=\frac{\text{dx}}{\text{x}}$
$\int\text{dy}=\int\frac{\text{dx}}{\text{x}}=\log_{\text{e}}\text{x+c}$
$\therefore\text{y}=\text{ln }\text{x + c}$
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MCQ 101 Mark
If $\frac{\text{dy}}{\text{dt}}=\text{ky}$ and $\text{k}\neq0$ which of the following could be the equation of $y:$
  • A
    $y = kx - 7$
  • ✓
    $y = 95e^{kt}$
  • C
    $y = 5 + ln k$
  • D
    $y = (x - k)^2$
Answer
Correct option: B.
$y = 95e^{kt}$

Given, $\frac{\text{dy}}{\text{dt}}=\text{ky}$ and $\text{k}\neq0$
We have $\frac{\text{dy}}{\text{y}}=\text{kdt}$
Apply integral on both sides, so we get $\int\frac{\text{dy}}{\text{y}}=\int\text{kdt}$
$\Rightarrow $ ln $(y) = kt + c ($where $c$ is constant$)$
$\Rightarrow y = e^ce^{kt} = ae^{kt} (a = e^c)$
The possible solution is $y = 95e^{kt}$

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Question 111 Mark
$\int_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}}\text{dx}$ is equal to:
  1. $2\sqrt{2}$
  2. $2(\sqrt{2+1})$
  3. $0$
  4. $2(\sqrt{2-1})$
Answer
  1. $0$
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Question 121 Mark
$\int(\text{x}-1)\text{e}^{-\text{x}}\text{ dx}$ is equal to:
  1. $-\text{x}\text{e}^{\text{x}}+\text{C}$
  2. $\text{x}\text{e}^{\text{x}}+\text{C}$
  3. $-\text{x}\text{e}^{-\text{x}}+\text{C}$
  4. $\text{x}\text{e}^{-\text{x}}+\text{C}$
Answer
  1. $-\text{x}\text{e}^{\text{x}}+\text{C}$
Solution:
$\int(\text{x}-1)\text{e}^{-\text{x}}\text{ dx}$
$=(\text{x}-1)\int\text{e}^{-\text{x}}\text{ dx}-\int\Big(\Big[\frac{\text{d}(\text{x}-1)}{\text{dx}}\Big]\int\text{e}^{-\text{x}}\text{dx}\Big)\text{ dx}$
$=(\text{x}-1)\frac{\text{e}^{-\text{x}}}{-1}-\int\frac{\text{e}^{-\text{x}}}{-1}\text{ dx}$
$=-(\text{x}-1)\text{e}^{-\text{x}}+\frac{\text{e}^{-\text{x}}}{-1}+\text{C}$
$=-\text{x}\text{e}^{-\text{x}}+\text{e}^{-\text{x}}+\text{C}$
$=-\text{x}\text{e}^{-\text{x}}+\text{C}$
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Question 131 Mark
$\int\frac{\sin^6\text{x}}{\cos^8\text{x}}\text{ dx}=$
  1. $\tan7\text{x}+\text{C}$
  2. $\frac{\tan^7\text{x}}{7}+\text{C}$
  3. $\frac{\tan7\text{x}}{7}+\text{C}$
  4. $\sec^7\text{x}+\text{C}$
Answer
  1. $\frac{\tan^7\text{x}}{7}+\text{C}$
Solution:
$\text{I}=\int\frac{\sin^6\text{x dx}}{\cos^8\text{x}}$
$\text{I}=\int\tan^6\text{x}\sec^2\text{x dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\text{t}^6\text{dt}$
$\text{I}=\frac{\text{t}^7}{7}+\text{C}$
$\text{I}=\frac{\tan^7\text{x}}{7}+\text{C}$
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Question 141 Mark
$\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^2(\text{x}\text{e}^{\text{x}})}\text{ dx}=$
  1. $2\log_\text{e}\cos(\text{xe}^{\text{x}})+\text{C}$
  2. $\sec(\text{xe}^{\text{x}})+\text{C}$
  3. $\tan(\text{xe}^{\text{x}})+\text{C}$
  4. $\tan(\text{x}+\text{e}^{\text{x}})+\text{C}$
Answer
  1. $\tan(\text{xe}^{\text{x}})+\text{C}$
Solution:
$\text{I}=\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^{2}(\text{xe}^{\text{x}})}\text{ dx}$
Put $\text{xe}^{\text{x}}=\text{t}$
$\text{e}^{\text{x}}(1+\text{x})\text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\cos^2\text{t}}$
$=\text{I}=\int\sec^2\text{t dt}$
$\text{t}=\tan\text{t}+\text{C}$
 $\text{I}=\tan(\text{xe}^{\text{x}})+\text{C}$
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Question 151 Mark
The symbol dx represents:
  1. The differential of the variable x
  2. The variable of integration is x
  3. Integral with respect to x
  4. None of the above
Answer
  1. The differential of the variable x
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Question 161 Mark
$\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{dx}$  is equal to:
  1. $2(\sin\text{x}+\text{x}\cos\theta)+\text{c}$
  2. $2(\sin\text{x}-\text{x}\cos\theta)+\text{c}$
  3. $2(\sin\text{x}+2\text{x}\cos\theta)+\text{c}$
  4. $2(\sin\text{x}-2\text{x}\cos\theta)+\text{c}$
Answer
  1. $2(\sin\text{x}+\text{x}\cos\theta)+\text{c}$
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Question 171 Mark
Solve: $\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$
  1. $\frac{1}{2}$
  2. $1$
  3. $2$
  4. $\frac{3}{2}$
Answer
  1. $2$
Solution:
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{xdx}}$
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$\int\limits_{0}^{\frac{\pi}{2}}(\sin\text{x}+\cos\text{x})\text{dx}$
$=[-\cos\text{x}+\sin\text{x}]\frac{\frac{\pi}{2}}{{0}}$
$=\Big[-\Big(\cos\frac{\pi}{2}-\cos0\Big)+\Big(\sin\frac{\pi}{2}-\sin0\Big)\Big]$
$=\big[-\big(0-1\big)+\big(1-0\big)\big]$
$=1+1=2$
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Question 181 Mark
The value of $\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{ dx}$ is:
  1. $\frac{\pi^2}{4}$
  2. $\frac{\pi^2}{2}$
  3. $\frac{3\pi^2}{2}$
  4. $\frac{\pi^2}{2}$
Answer
  1. $8$
Solution:
$\int\limits^{2\pi}_0\sqrt{1+\sin\frac{\text{x}}{2}}\text{dx}$
$=\int\limits^{2\pi}_0\sqrt{\sin^2\frac{\text{x}}{4}+\cos^2\frac{\text{x}}{4}+2\sin\frac{\text{x}}{4}\cos\frac{\text{x}}{4}}\text{dx}$
$=\int\limits^{2\pi}_0\Big(\sin\frac{\text{x}}{4}+\cos\frac{\text{x}}{4}\Big)\text{dx}$
$=\Bigg[\frac{-\cos^\frac{\text{x}}{4}}{\frac{1}{4}}+\frac{\sin\frac{\text{x}}{4}}{\frac{1}{4}}\Bigg]^{2\pi}_0$
$=4\Big[\frac{\text{x}}{4}-\cos\frac{\text{x}}{4}\Big]^{2\pi}_0$
$=4\Big[\sin\frac{2\pi}{4}-\cos\frac{2\pi}{4}-\sin0+\cos0\Big]$
$=4\Big[\sin\frac{\pi}{2}-\cos\frac{\pi}{2}-0+1\Big]$
$=4\big[1-0-0+1\big]$
$=4\times2$
$=8$
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Question 191 Mark
$\int\frac{\text{dx}}{1+\text{cos x}}=$
  1. $\tan\frac{\text{x}}{2}+\text{k}$
  2. $\frac{1}{2}\tan\frac{\text{x}}{2}+\text{k}$
  3. $2\tan\frac{\text{x}}{2}+\text{k}$
  4. $\tan^2\frac{\text{x}}{2}+\text{k}$
Answer
  1. $\tan\frac{\text{x}}{2}+\text{k}$
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Question 201 Mark
The antiderivative of every odd function is:
  1. An odd function
  2. An even function
  3. Neither even nor odd
  4. Sometimes even, sometimes odd
Answer
  1. An even function
Solution:
The anti derivative of an odd function is even. Let f(x) be odd
eg = f(x) = x odd function
$\int\text{xdx}=\frac{\text{x}^2}{2}+\text{c}$
$\text{g}'(\text{x})=\frac{{\text{x}}^{2}}{\text{x}}+\text{c}$ is even.
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Question 211 Mark
$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{ dx}$ is equal to:
  1. $\frac{\pi}{12}$
  2. $\frac{\pi}{6}$
  3. $\frac{\pi}{4}$
  4. $\frac{\pi}{3}$
Answer
  1. $\frac{\pi}{12}$
Solution:
$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{dx}$
$=\big[\tan^{-1}\text{x}\big]^\sqrt{3}_1$
$=\frac{\pi}{3}-\frac{\pi}{4}$
$=\frac{\pi}{12}$
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Question 221 Mark
The value of the integral $\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}$ is:
  1. $0$
  2. $\frac{\pi}{2}$
  3. $\frac{\pi}{4}$
  4. none of these.
Answer
  1. $\frac{\pi}{4}$
Solution:
$\text{ I}=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\cos\text{x}}}{\sqrt\cos\text{x}+\sqrt{\sin\text{x}}}\text{dx}\ ...(\text{i})$
$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)+\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}}\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\sin{\text{x}}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{dx}...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^\frac{\pi}{2}_0\bigg[\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}+\frac{\sqrt{\sin\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\bigg]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\text{dx}$
$=[\text{x}]^\frac{\pi}{2}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
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Question 231 Mark
If $\int\text{f(x)}\text{dx}=2\text{ f(x)}^3+\text{c}$ and $\text{f(x)}\neq0$ then f(x) is:
  1. $\frac{\text{x}}{2}$
  2. $\text{x}^3$
  3. $\frac{1}{\sqrt{\text{x}}}$
  4. $\sqrt{\frac{\text{x}}{3}}$
Answer
  1. $\sqrt{\frac{\text{x}}{3}}$
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MCQ 241 Mark
If $\frac{\text{dy}}{\text{dx}}=\text{x}^{-3}$ then $y:$
  • ✓
    $\frac{-1}{2\text{x}^{2}}+\text{c}$
  • B
    $\frac{-\text{x}^{-4}}{4}+\text{c}$
  • C
    $\frac{2}{\text{x}^{2}}+\text{c}$
  • D
    $\frac{\text{x}^{-2}}{2}+\text{c}$
Answer
Correct option: A.
$\frac{-1}{2\text{x}^{2}}+\text{c}$

we have $\frac{\text{dy}}{\text{dx}}=\text{x}^{-3}$
$\Rightarrow dy = x^{-3} dx$
Integrating both sides 
$\therefore\text{y}=\int\text{x}^{-3}\text{dx}=\frac{\text{x}^{-3 + 1}}{-3+1}+\text{c}=-\frac{1}{2\text{x}^2}+\text{c}$

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Question 251 Mark
$\int_{0}^{1}\text{x}(1-\text{x})^{99}$ is equal to:
  1. $\frac{1}{10010}$
  2. $\frac{1}{10100}$
  3. $\frac{1}{1010}$
  4. $\frac{11}{10100}$
Answer
  1. $\frac{1}{10100}$
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Question 261 Mark
$\int\frac{9\text{x}}{9\text{x}^2+1}=$
  1. $\frac{1}{3}\tan^{-1}(2\text{x})+\text{C}$
  2. $\frac{1}{3}\tan^{-1}\text{x}+\text{C}$
  3. $\frac{1}{3}\tan^{-1}(3\text{x})+\text{C}$
  4. $\frac{1}{3}\tan^{-1}(6\text{x})+\text{C}$
Answer
  1. $\frac{1}{3}\tan^{-1}(3\text{x})+\text{C}$
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Question 271 Mark
Choose the correct answer in Exercise:
$\int\text{e}^\text{x}\sec\text{x}(1+\tan\text{x})\text{dx}$ equals
  1. $\text{e}^\text{x}\cos\text{x}+\text{C}$
  2. $\text{e}^\text{x}\sec\text{x}+\text{C}$
  3. $\text{e}^\text{x}\sin\text{x}+\text{C}$
  4. $\text{e}^\text{x}\tan\text{x}+\text{C}$
Answer
  1. $\text{e}^\text{x}\sec\text{x}+\text{C}$
$\int\text{e}^\text{x}\sec\text{x}(1+\tan\text{x})\text{dx}$
Let $\text{I}=\int\text{e}^\text{x}\sec\text{x}(1+\tan\text{x})\text{dx}=\int\text{e}^\text{x}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}​​$
Also, let $\sec\text{x}=\text{f}(\text{x})\Rightarrow \ \sec\text{x}\tan\text{x}=\text{f}'(\text{x})$
It is known that, $\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}=\text{e}^\text{x}\text{f}(\text{x})+\text{C}$
$\therefore\ \text{I}=\text{e}^\text{x}\sec\text{x}+\text{C}$​​​​​​​
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Question 281 Mark
$\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+2\sin\text{x}}}\text{dx}=$
  1. $\log(\sin\text{x}-\cos\text{x})$
  2. $\text{x}$
  3. $\log\text{x}$
  4. $\log\sin(\cos\text{x})$
Answer
  1. $\text{x}$
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Question 291 Mark
$\int\frac{\sin\text{x} + \cos\text{x}}{\sqrt{(1 + \sin2\text{x})}}\text{dx}$ is:
  1. $\sin\text{x + c}$
  2. $\text{x + c}$
  3. $\cos\text{x + c}$
  4. $\tan\text{x + c}$
Answer
  1. $\text{x + c}$
Solution:
$\int\frac{\sin\text{x} + \cos\text{x}}{\sqrt{(1 + \sin2\text{x})}}\text{dx}$
$\int\frac{\sin\text{x} + \cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{dx}$
$=\int1\text{dx}$
$=\text{x + c}$
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Question 301 Mark
Choose the correct option from given four options:
If $\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C},$ then:
  1. $\text{a}=\frac{-1}{10},\text{b}=\frac{-2}{5}$
  2. $\text{a}=\frac{-1}{10},\text{b}=-\frac{2}{5}$
  3. $\text{a}=\frac{-1}{10},\text{b}=\frac{2}{5}$
  4. $\text{a}=\frac{1}{10},\text{b}=\frac{2}{5}$
Answer
  1. $\text{a}=\frac{-1}{10},\text{b}=\frac{2}{5}$
Solution:
Given that, $\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C}$
Now, $\text{I}=\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}$
$\frac{1}{(\text{x}+2)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow1=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+2)$
$\Rightarrow1=\text{Ax}^2+\text{A}+\text{Bx}^2+2\text{Bx}+\text{Cx}+2\text{C}$
$\Rightarrow1=(\text{A}+\text{B})\text{x}^2+(2\text{B}+\text{C})\text{x}+\text{A}+2\text{C}$
$\Rightarrow\text{A}+\text{B}=0,\text{A}+2\text{C}=1,2\text{B}+\text{C}=0$
We have, $\text{A}=\frac{1}{5},\text{B}=-\frac{1}{5}$ and $\text{C}=\frac{2}{5}$
$\therefore\ \int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\frac{1}{5}\int\frac{1}{\text{x}+2}\text{dx}+\int\frac{-\frac{1}{5}\text{x}+\frac{2}{5}}{\text{x}^2+1}\text{dx}$
$=\frac{1}{5}\int\frac{1}{\text{x}+2}\text{dx}-\frac{1}{5}\int\frac{\text{x}}{1+\text{x}^2}\text{dx}+\frac{1}{5}\int\frac{2}{1+\text{x}^2}\text{dx}$
$=\frac{1}{5}\log|\text{x}+2|-\frac{1}{10}\log|1+\text{x}^2|+\frac{2}{5}\tan^{-1}\text{x}+\text{C}$
$\therefore\ \text{b}=\frac{2}{5}$ and $\text{a}=\frac{-1}{10}$
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Question 311 Mark
The value of $\int\limits^{\pi}_0\frac{1}{5+3\cos\text{x}}\text{ dx}$ is:
  1. $\frac{\pi}{4}$
  2. $\frac{\pi}{8}$
  3. $\frac{\pi}{2}$
  4. $0$
Answer
  1. $\frac{\pi}{4}$
Solution:
$\int\limits^{\pi}_0\frac{1}{5+3\cos\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{1}{5+3\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{5+5\tan^{2}\frac{\text{x}}{2}+3-3\tan^{2}\frac{\text{x}}{2}}$
$=\int\limits^{\pi}_0\frac{\sec^2\frac{\text{x}}{2}}{8+2\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t},$ then $\sec^2\frac{\text{x}}{2}\text{ dx}=2\text{dt}$
When $\text{x}=0,\text{ t}=0,\text{x}=\pi,\text{ t}=\infty$
Therefore the integral becomes
$\frac{1}{2}\int\limits^{\infty}_0\frac{\text{dt}}{4+\text{t}^2}$
$=\frac{1}{2}\Big[\tan^{-1}\frac{\text{t}}{2}\Big]^{\infty}_0$
$=\frac{1}{2}\Big(\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{4}$
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Question 321 Mark
$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$ is equal to:
  1. $\tan\text{x}+\cos\text{x}+\text{c}$
  2. $\tan\text{x}+\text{cosec}\text{x}+\text{c}$
  3. $\tan\text{x}+\text{cot}\text{x}+\text{c}$
  4. $\tan\text{x}+\sec\text{x}+\text{c}$
Answer
  1. $\tan\text{x}+\text{cot}\text{x}+\text{c}$
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Question 331 Mark
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$ is equal to:
  1. 1
  2. 2
  3. -1
  4. -2
Answer
  1. $2$
Solution:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$
$=-\int\limits^0_{-\frac{\pi}{2}}\sin\text{x}\text{ dx}+\int\limits^\frac{\pi}{2}_0\sin\text{x}\text{ dx}$
$=-\big[-\cos\text{x}\big]^0_{-\frac{\pi}{2}}+\big[-\cos\text{x}\big]^\frac{\pi}{2}_0$
$=1-0-0+1$
$=2$
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Question 341 Mark
$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....+\frac{1}{2\text{n}+\text{n}}\Big\}$ is equal to:
  1. $\ln\Big(\frac{1}{3}\Big)$
  2. $\ln\Big(\frac{2}{3}\Big)$
  3. $\ln\Big(\frac{3}{2}\Big)$
  4. $\ln\Big(\frac{4}{3}\Big)$
Answer
  1.  $\ln\Big(\frac{3}{2}\Big)$
Solution:
$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....\ +\frac{1}{2\text{n}+\text{n}}\Big\}$
$=\lim\limits_{\text{n}\rightarrow\infty}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2\text{n}+\text{r}}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2+\frac{\text{r}}{\text{n}}}$
let $\frac{\text{r}}{\text{n}}=\text{x}$
$=\int\limits^\infty_0\frac{1}{2+\text{x}}\text{dx}$
$=\Big[\log\big(2+\text{x}\big)\Big]^\infty_0$
$=\log3-\log2$
$=\log\frac{3}{2}$
$=\ln\Big(\frac{3}{2}\Big)$
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Question 351 Mark
Evaluate: $\int\frac{3\text{x}^2+1}{(\text{x}^2-1)^3}\text{dx}$
  1. $\text{c}-\frac{\text{x}}{(\text{x}^2-1)^2}$
  2. $\text{c}-\frac{\text{x}}{(\text{x}^2+1)^2}$
  3. $\frac{\text{x}}{(\text{x}^2-1)^2}$
  4. $\frac{\text{x}}{(\text{x}^2+1)^2}$
Answer
  1. $\text{c}-\frac{\text{x}}{(\text{x}^2-1)^2}$
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Question 361 Mark
$\int\cos(\log_\text{e}.\text{x})\text{dx}$ is equal to:
  1. $\frac{1}{2}\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
  2. $\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
  3. $\frac{1}{2}\text{x}[\cos(\log_\text{e}\text{x})-\sin(\log_\text{e}\text{x})]$
  4. $\text{x}[\cos(\log_\text{e}\text{x})-\sin(\log_\text{e}\text{x})]$
Answer
  1. $\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
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Question 371 Mark
$\int\frac{\text{dx}}{1-\cos\text{x}-\sin\text{x}}$ is equal to:
  1. $\log|1+\cot\frac{\text{x}}{2}|+\text{c}$
  2. $\log|1-\tan\frac{\text{x}}{2}|+\text{c}$
  3. $\log|1-\cot\frac{\text{x}}{2}|+\text{c}$
  4. $\log|1+\tan\frac{\text{x}}{2}|+\text{c}$
Answer
  1. $\log|1-\cot\frac{\text{x}}{2}|+\text{c}$
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Question 381 Mark
$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{\text{dx}}{1+\cos2\text{x}}\text{dx}$ is equal to:
  1. 1
  2. 2
  3. 3
  4. 4
Answer
  1. 1
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Question 391 Mark
Calculate: $\int(\text{x}^3-\frac{1}{\text{x}}+{3\text{x}})\text{dx:}$
  1. $\frac{\text{x}^{4}}{3}-\log\text{x}+\frac{\text{5x}^{2}}{2}+\text{c}$
  2. $\frac{\text{x}^{4}}{4}-\log\text{x}+\frac{\text{2x}^{2}}{3}+\text{c}$
  3. $\frac{\text{x}^{4}}{4}-\log\text{x}+\frac{\text{3x}^{2}}{4}+\text{c}$
  4. $\frac{\text{x}^{4}}{4}-\log\text{x}+\frac{\text{3x}^{2}}{2}+\text{c}$
Answer
  1. $\frac{\text{x}^{4}}{4}-\log\text{x}+\frac{\text{3x}^{2}}{2}+\text{c}$
Solution:
$\int(\text{x}^3-\frac{1}{\text{x}}+{3\text{x}})\text{dx}=\frac{x^4}{4}-\text{ln}\mid{\text{x}}\mid{+}\frac{3\text{x}^2}{2}+\text{c}$
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Question 401 Mark
$\int\text{e}^\text{x}(\frac{1-\text{x}}{1+\text{x}^2})^2\text{dx}$ is equal to:
  1. $\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
  2. $-\frac{-\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
  3. $\frac{\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{c}$
  4. $\frac{-\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{c}$
Answer
  1. $\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
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Question 411 Mark
Evaluate $\int\cos^3\text{xe}^{\log{\sin}\text{x}}\text{dx}$
  1. $-\frac{\cos^4\text{x}}{4}+\text{C}$
  2. $-\frac{\sin\text{x}}{\text{x}^2}+\text{C}$
  3. $-\frac{\cos^3\text{x}}{3}+\text{C}$
  4. None of these 
Answer
  1. $-\frac{\cos^4\text{x}}{4}+\text{C}$
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Question 421 Mark
What is $\int\frac{\text{dx}}{\text{x}(1+\text{lnx})^\text{n}}$ equal to $(\text{n}\neq1)$
  1. $\frac{{1}}{{(\text{n}-1)}(1+\text{lnx})^{\text{n}-1}}+\text{c}$
  2. $\frac{1-\text{n}}{(1+\text{lnx})^{1-\text{n}}}+\text{c}$
  3. $\frac{{\text{n}+1}}{{(1+\text{lnx})}^{\text{n}+1}}+\text{c}$
  4. $-\frac{1}{(\text{n}+1)(1+\text{lnx})^{\text{n}-1}}+\text{c}$
Answer
  1. $-\frac{1}{(\text{n}+1)(1+\text{lnx})^{\text{n}-1}}+\text{c}$
Solution:
Given,
$\int\frac{1}{\text{x}(1+\text{ln}(\text{x}))^\text{n}}\text{dx}$
apply u = 1 + ln (x)
$=\int\frac{1}{\text{u}^{\text{n}}}\text{du}$
$=\int\text{u}^{-\text{n}}\text{du}$
$=\frac{\text{u}^{-\text{n}+1}}{-\text{n}+1}$
$=\frac{(1+\text{ln}(\text{x}))^{-{\text{n}+1}}}{{-\text{n}+1}}$
$-\frac{1}{(\text{n}+1)(1+\text{lnx})^{\text{n}-1}}+\text{c}$
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Question 431 Mark
$\int\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}=$
  1. $-\text{e}^{\text{x}}\tan\frac{\text{x}}{2}+\text{C}$
  2. $-\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$
  3. $-\frac{1}{2}\text{e}^{\text{x}}\tan\frac{\text{x}}{2}+\text{C}$
  4. $-\frac{1}{2}\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$
Answer
  1. $-\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$
Solution:
Let $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
$\int\text{e}^{\text{x}}\Big(\frac{1}{1-\cos\text{x}}-\frac{\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
$\int\text{e}^{\text{x}}\bigg(\frac{1}{2\sin^2\frac{\text{x}}{2}}-\frac{2\sin\frac{\pi}{2}\cos\frac{\pi}{2}}{2\sin^2\frac{\pi}{2}}\bigg)\text{dx}$
$\int\text{e}^\text{x}\Big(\frac{1}{2}\text{cosec}^2\frac{\text{x}}{2}-\cot\frac{\text{x}}{2}\Big)\text{dx}$
As, we know that $\int\text{e}^{\text{x}}\{\text{f(x)}+\text{f}'(\text{x})\}\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$
$\therefore\ \text{I}=-\text{e}^{\text{x}}\cot\Big(\frac{\text{x}}{2}\Big)+\text{C}$
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Question 441 Mark
For $\text{x, }\epsilon\text{ R},\text{f}(\text{x})=\mid\log2-\sin\text{x}\mid$ and g(x) = f(f(x)) then:
  1. $\text{g}'(0)=\cos (\log2)$
  2. $\text{g}'(0)=-\cos (\log2)$
  3. $\text{g }\text{is diffrerentible at x = 0 and }\text{g}'(0)=-\sin(\log2)$
  4. $\text{g }\text{is diffrerentible at x = 0 }$
Answer
  1. $\text{g}'(0)=\cos (\log2)$
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Question 451 Mark
Choose the correct option from given four options:
$\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{dx}$ is equal to:
  1. $2(\sin+\text{x}\cos\theta)+\text{C}$
  2. $2(\sin-\text{x}\cos\theta)+\text{C}$
  3. $2(\sin+2\text{x}\cos\theta)+\text{C}$
  4. $2(\sin-2\text{x}\cos\theta)+\text{C}$
Answer
  1. $2(\sin+\text{x}\cos\theta)+\text{C}$
Solution:
Let $\text{I}=\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{dx}$
$=\int\frac{\big(2\cos^2\text{x}-1-2\cos^2\theta+1)}{\cos\text{x}-\cos\theta}\text{dx}$
$=2\int\frac{\big(\cos\text{x}+\cos\theta)(\cos\text{x}-\cos\theta)}{\cos\text{x}-\cos\theta}\text{dx}$
$=2\int(\cos\text{x}+\cos\theta)\text{dx}$
$=2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$
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Question 461 Mark
If $\text{g}'(\text{x})=\int\text{x}^\text{x}\log_\text{e}(\text{ex})\text{dx}$ then $\text{g}(\pi)$ equals:
  1. $\pi\log_\text{e}\pi$
  2. $\pi^\pi\log_\text{e}(\text{e}\pi)$
  3. $\pi^\pi\log_\text{e}(\pi)$
  4. $\pi^\pi$
Answer
  1. $\pi^\pi$
Solution:
$\text{g}'(\text{x})=\int\text{x}^{\text{x}}(1+\log{\text{e}^\text{x}})\text{dx}$
$=\int\text{d}(\text{x}^{\text{x}})$
$\text{g'}(\text{x})=\text{x}^{\text{x}}$
$\text{g'}({\pi})={\pi}^{\pi}$
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Question 471 Mark
If $(\text{a}+\text{b}-\text{x})=\text{f}(\text{x}),$ then $\int\limits^\text{b}_\text{a}\text{x f}(\text{x})\text{dx}$ is equal to:
  1. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{b}-\text{x})\text{dx}$
  2. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{b}+\text{x})\text{dx}$
  3. $\frac{\text{b}-\text{a}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$
  4. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$
Answer
  1. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$
Solution:
Let, $\text{I}=\int\limits^\text{b}_\text{a}\text{x}\text{ f}(\text{x})\text{dx}\ ....(\text{i})$
$=\int\limits^\text{b}_\text{a}(\text{a}+\text{b}-\text{x})\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$
$=\int\limits^\text{b}_\text{a}(\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^\text{b}_\text{a}(\text{x}+\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}$
$=(\text{a}+\text{b})\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$
Hence $\text{I}=\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$
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Question 491 Mark
$\int\frac{1}{7+5\cos\text{x}}\text{ dx}=$
  1. $\frac{1}{\sqrt{6}}\tan^{-1}\Big(\frac{1}{\sqrt{6}}\tan\frac{\text{x}}{2}\Big)+\text{C}$
  2. $\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\tan\frac{\text{x}}{2}\Big)+\text{C}$
  3. $\frac{1}{4}\tan^{-1}\Big(\tan\frac{\text{x}}{2}\Big)+\text{C}$
  4. $\frac{1}{7}\tan^{-1}\Big(\tan\frac{\text{x}}{2}\Big)+\text{C}$
Answer
  1. $\frac{1}{\sqrt{6}}\tan^{-1}\Big(\frac{1}{\sqrt{6}}\tan\frac{\text{x}}{2}\Big)+\text{C}$
Solution:
Let $\text{I}=\int\frac{\text{dx}}{7+5\cos\text{x}}$
Putting $\cos\text{x}=\frac{1-\tan^2\frac{\pi}{2}}{1+\tan^2\frac{\pi}{2}}$
$\therefore\ \text{I}=\int\frac{\text{dx}}{7+5\times\bigg(\frac{1-\tan^2\frac{\pi}{2}}{1+\tan^2\frac{\pi}{2}}\bigg)}$
$=\int\frac{\big(1+\tan^2\frac{\pi}{2}\big)\text{dx}}{7\big(1+\tan^2\frac{\pi}{2}\big)+5-5\tan^2\frac{\pi}{2}}$
$=\int\frac{\sec^2\frac{\pi}{2}\text{dx}}{2\tan^2\frac{\pi}{2}+12}$
$=\frac{1}{2}\int\frac{\sec^2\frac{\pi}{2}\text{dx}}{\tan^2\frac{\pi}{2}+(\sqrt{6})^2}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{dt}$
$\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=2\text{dt}$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{2\text{ dt}}{\text{t}^2+(\sqrt{6})^2}$
$=\frac{1}{\sqrt{6}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{6}}\Big)+\text{C}$ $\Big(\because\int\frac{1}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}+\text{C}\Big)$
$=\frac{1}{\sqrt{6}}\tan^{-1}\bigg(\frac{\tan\frac{\pi}{2}}{\sqrt{16}}\bigg)+\text{C}$ $\Big(\because\text{t}=\tan\frac{\text{x}}{2}\Big)$
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Question 501 Mark
Choose the correct answer in Exercise:
$\int\frac{\cos2\text{x}}{(\sin\text{x}+\cos\text{x)}^{2}}$ is equal to
  1. $\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
  2. $\log|\sin\text{x}+\cos\text{x}|+\text{C}$
  3. $\log|\sin\text{x}-\cos\text{x}|+\text{C}$
  4. $\frac{1}{(\sin\text{x}+\cos\text{x)}^{2}}+\text{C}$
Answer
  1. $\log|\sin\text{x}+\cos\text{x}|+\text{C}$ 
$\text{Let I}=\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x)}^{2}}$
$\text{I}=\int\frac{(\cos^{2}\text{x}-\sin^{2}\text{x)}}{(\cos\text{x}+\sin\text{x)}^{2}}\text{dx}$
$=\int\frac{(\cos\text{x}+\sin\text{x})(\cos\text{x}-\sin\text{x)}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$
$\text{Let}\cos\text{x}+\sin\text{x}=\text{t}\Rightarrow(\cos\text{x}-\sin\text{x)}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\cos\text{x}+\sin\text{x}|+\text{C}$
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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip