M.C.Q (1 Marks) - Page 3 - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

Question 1011 Mark

If $\int\frac{\sin^8\text{x}-\cos^8\text{x}}{1-2\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\text{a}\sin2\text{x}+\text{C},$ then a =

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

Answer

$-\frac{1}{2}$

Solution:
$\int\frac{\sin^8\text{x}-\cos^8\text{x}}{1-2\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\text{a}\sin2\text{x}+\text{C}\ ...(\text{i})$
Considering LHS of eq. (i)
$\Rightarrow\int\frac{(\sin^4\text{x}-\cos^4\text{x})(\sin^4\text{x}+\cos^4\text{x})}{(1-2\sin^2\text{x}\cos^2\text{x})}$
$\Rightarrow\frac{(\sin^2\text{x}-\cos^2\text{x})(\sin^2\text{x}+\cos^2\text{x})\cdot(\sin^4\text{x}+\cos^4\text{x})\text{ dx}}{\big\{(\sin^2\text{x}+\cos^2\text{x})^2-2\sin^2\text{x}\cos^2\text{x}\big\}}$
$\Rightarrow\int\frac{(\sin^2\text{x}-\cos^2\text{x})\cdot(\sin^4\text{x}+\cos^4\text{x})\text{ dx}}{(\sin^4\text{x}+\cos^4\text{x}+2\sin^2\text{x}\cos^2\text{x}-2\sin^2\text{x}\cos^2\text{x})}$
$\Rightarrow-\int\frac{(\cos^2\text{x}-\sin^2\text{x})\times(\sin^4\text{x}+\cos^4\text{x})\text{ dx}}{(\sin^4\text{x}+\cos^4\text{x})}$
$\Rightarrow-\int\cos(2\text{x})\text{ dx}\ ...(\text{ii})$ $(\because\cos^2\text{x}-\sin^2\text{x}=\cos2\text{x})$
Comparing the RHS of eq. (i) with eq. (ii) we get
$\text{a}=-\frac{1}{2}$

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Question 1021 Mark

$\int\limits^\frac{\pi}{2}_0\frac{1}{1+\tan\text{x}}\text{dx}$ is equal to:

$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\pi$

Answer

$\frac{\pi}{4}$

Solution:
Let, $\text{I}=\int\limits^\frac{\pi}{2}_0\frac{1}{1+\tan\text{x}}\text{dx}\ ...(\text{i})$
$=\int\limits^\frac{\pi}{2}_0\frac{1}{1+\tan\big(\frac{\pi}{2}-\text{x}\big)}\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{1}{1+\cot\text{x}}\text{dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^\frac{\pi}{2}_0\Big[\frac{1}{1+\tan\text{x}}+\frac{1}{1+\cot\text{x}}\Big]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\bigg[\frac{(1+\cot\text{x})+(1+\tan\text{x})}{(1+\tan\text{x})(1+\cot\text{x})}\bigg]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\Big[\frac{2+\tan\text{x}+\cot\text{x}}{1+\tan\text{x}+\cot\text{x}+\tan\text{x}\cot\text{x}}\Big]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\Big[\frac{2+\tan\text{x}+\cot\text{x}}{2+\tan\text{x}+\cot\text{x}}\Big]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\text{dx}$
$=\big[\text{x}\big]^\frac{\pi}{2}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$

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Question 1031 Mark

$\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\cos\text{x}}\text{ dx}=$

$\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$
$\frac{\pi}{\text{ab}}$
$\frac{\pi}{\text{a}^2-\text{b}^2}$
$({\text{a}+\text{b}})\pi$

Answer

$\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$

Solution:
We have,
$\text{I}=\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\cos\text{x}}\text{ dx}$
$=\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$
$=\int\limits^\pi_0\frac{1+\tan^2\frac{\text{x}}{2}}{\text{a}\Big(1+\tan^2\frac{\text{x}}{2}\Big)+\Big(1-\tan^2\frac{\text{x}}{2}\Big)}\text{ dx}$
$=\int\limits^\pi_0\frac{1+\tan^2\frac{\text{x}}{2}}{(\text{a}+\text{b})+(\text{a}-\text{b})\tan^2\frac{\text{x}}{2}}\text{ dx}$
$=\int\limits^\pi_0\frac{\sec^2\frac{\text{x}}{2}}{(\text{a}+\text{b})+(\text{a}-\text{b})\tan^2\frac{\text{x}}{2}}\text{ dx}$
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 $\text{x}\rightarrow\pi;\text{ t}\rightarrow\infty$
$\therefore\ \text{I}=\int\limits^\pi_0\frac{2\text{dt}}{(\text{a}+\text{b})+(\text{a}-\text{b})\text{t}^2}$
$=\frac{2}{\text{a}-\text{b}}\int\limits^\pi_0\frac{1}{\big(\frac{\text{a}+\text{b}}{\text{a}-\text{b}}\big)+\text{t}^2}\text{dt}$
$=\frac{2}{(\text{a}-\text{b})}\int\limits^\infty_0\frac{1}{\Big(\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}\Big)^2+\text{t}^2}\text{ dt}$
$=\frac{2}{(\text{a}-\text{b})}\times\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$ $\Bigg[\tan^{-1}\frac{\text{t}}{\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}}\Bigg]^\infty_0$
$=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\Big[\frac{\pi}{2}\Big]$
$=\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$

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Question 1041 Mark

Evaluate: $\int\frac{1}{\sqrt{9+8\text{x}-\text{x}^2}}\text{dx}.$

$-\sin^{-1}(\frac{\text{x}-4}{5})+\text{c}$
$\sin^{-1}(\frac{\text{x}+4}{5})+\text{c}$
$\sin^{-1}(\frac{\text{x}-4}{5})+\text{c}$
$\text{None of there}$

Answer

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

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Question 1051 Mark

Evaluate: $\int\frac{1}{\sqrt{1-\text{e}^{\text{2x}}}}\text{dx}.$

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

Answer

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

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Question 1061 Mark

Choose the correct answer in Exercise. $\int\sqrt{\text{x}^2-8\text{x}+7}\text{dx}$ is equal to

$\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}+9\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
$\frac{1}{2}(\text{x}+4)\sqrt{\text{x}^2-8\text{x}+7}+9\text{log}\Bigg|\text{x}+4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
$\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}-3\sqrt2\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
$\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}-\frac{9}{2}\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$

Answer

$\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}+9\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$

$=\Bigg(\frac{\text{x}-4}{2}\Bigg)\sqrt{(\text{x}-4)^2-3^2}-\frac{3^2}{2}\text{log}\Big|\text{x}-4+\sqrt{(\text{x}-4)^2-3^2}\Big|+\text{C}$
$\Bigg[\therefore\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|\Bigg]$
$=\Bigg(\frac{\text{x}-4}{2}\Bigg)\sqrt{\text{x}^2-8\text{x}+7}-\frac{9}{2}\text{log}\Big|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Big|+\text{C}$

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Question 1071 Mark

$\int\frac{\text{dx}}{\sqrt{\text{x}}}=$

$\sqrt{\text{x}}+\text{k}$
$2\sqrt{\text{x}}+\text{k}$
$\text{x}+\text{k}$
$\frac{2}{3}\times\frac{3}{2}+\text{k}$

Answer

$2\sqrt{\text{x}}+\text{k}$

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Question 1081 Mark

$\int\frac{\text{x}}{4+\text{x}^4}\text{ dx}$ is equal to:

$\frac{1}{4}\tan^{-1}\text{x}^2+\text{C}$
$\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
$\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
none of these.

Answer

$\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$

Solution:
$\text{I}=\int\frac{\text{x}}{4+\text{x}^4}\text{ dx}$
Put $\text{x}^2=\text{t}$
$2\text{xdx}=\text{dt}$
$\text{x dx}=\frac{\text{dt}}{2}$
$\text{I}=\int\frac{\frac{\text{dt}}{2}}{4+\text{t}^2}$
$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$
$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)+\text{C}$

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MCQ 1091 Mark

$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}=$

✓
$\pi\ln 2$
B
$-\pi\ln2$
C
$0$
D
$-\frac{\pi}{2}\ln2$

Answer

Correct option: A.

$\pi\ln 2$

$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}$
Substitute $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{d}\theta$
When,
$\text{x}=0\Rightarrow\theta=0$
$\text{x}=\infty\Rightarrow\theta=\frac{\pi}{2}$
$\int\limits^{\frac{\pi}{2}}_0\Big(\tan\theta+\frac{1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}}\times\sec^2\theta\text{d}\theta$
$\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\tan^2\theta+1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}\theta}\times\sec^2\theta\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\frac{1}{\sec^2\theta}\times\sec^2\theta\text{d}\theta$ $\big[\because1+\tan^2\theta=\sec^2\theta\big]$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{1}{\sin\theta\cdot\cos\theta}\Big)\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\big(\sin\theta\cos\theta\big)\text{d}\theta$
$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\big[\log\sin\theta+\log\cos\theta\big]\text{d}\theta$
$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta$
Let us conside,
$\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta=\text{I}\ ....(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big(\sin\Big(\frac{\pi}{2}-\theta\Big)\Big)\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta\ ...(\text{ii})$
Adding $(i)$ and $(ii)$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta+\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$
$=\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta\cdot\cos\theta)\text{d}\theta$
$=\int\limits^{\frac{\pi}{2}}_0\log\big(\sin2\theta\big)\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log2\text{ d}\theta$
Let us consider $2\theta=\text{t}$
$2\text{d}\theta=\text{dt}$
$2\text{I}=\frac{1}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$
$2\text{I}=\frac{2}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$ $\big[\because\sin\theta$ is positive in both $1^{st}$ and $2^{nd} $ quadrants$\big]$
$2\text{I}=\text{I}-\frac{\pi}{2}\log2$
$2\text{I}-\text{I}=-\frac{\pi}{2}\log2$
$\text{I}=-\frac{\pi}{2}\log2,$ where $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta$
Now,
$-\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta)\text{ d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$
$-2\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta=-2\times\text{I}$
$=-2\times-\frac{\pi}{2}\log2$ $\Big[\text{Where I}=-\frac{\pi}{2}\log2\Big]$
$=\pi\log2$

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Question 1101 Mark

$\int\frac{\text{x}^3}{\text{x}+1}\text{ dx}$ 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:
$\text{I}=\int\frac{\text{x}^3}{\text{x}+1}\text{ dx}$
$\text{I}=\int\frac{\text{x}^3+1-1}{\text{x}+1}\text{ dx}$
$\text{I}=\int\frac{(\text{x}+1)(\text{x}^2-\text{x}+1)}{\text{x}+1}\text{ dx}$
$\text{I}=\int\Big(\text{x}^2-\text{x}+1-\frac{1}{\text{x}+1}\Big)\text{dx}$
$\text{I}=\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$

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Question 1111 Mark

Evaluate the following integral: $\int\sec^2\text{xdx:}$

$2\tan\text{x+c}$
$\tan2\text{x}+\text{c}$
$\tan\text{x}+\text{c}$
None of these

Answer

$\tan\text{x}+\text{c}$

Solution:
We know that. $\frac{\text{d}}{\text{dx}}(\tan\text{x})=\sec^2\text{x}$
$\therefore\text{d }(\tan\text{x})=\sec^2\text{xdx}$
$\Rightarrow\text{I}=\int\sec^2\text{xdx}=\tan\text{x+c}$

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Question 1121 Mark

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:

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

Answer

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

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Question 1131 Mark

If $\int\frac{2^{\frac{1}{\text{x}}}}{\text{x}^2}\text{dx}=\text{k }2^{\frac{1}{\text{x}}}+\text{C},$ then k is equal to:

$-\frac{1}{\log_\text{e}2}$
$-\log_\text{e}2$
$-1$
$\frac{1}{2}$

Answer

$-\frac{1}{\log_\text{e}2}$

Solution:
$\text{k}=\int\frac{2^{\frac{1}{\text{x}}}}{\text{x}^2}\text{ dx}$
Put $\frac{1}{\text{x}}=\text{t}$
$\frac{-1}{\text{x}^2}\text{ dx}=\text{dt}$
$\frac{1}{\text{x}^2}\text{ dx}=-\text{dt}$
$\text{k}=\int2^{\text{t}}(-\text{dt})$
$\text{k}=\frac{-2^{\text{t}}}{\log_\text{e}2}+\text{C}$
$\text{k}=\frac{-2^{\frac{1}{\text{x}}}}{\log_\text{e}2}+\text{C}$
$\text{k}=\frac{-1}{\log_\text{e}2}$

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Question 1141 Mark

$\int\frac{\text{x}^3}{\text{x}+1}$ 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

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

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Question 1151 Mark

If $\frac{\text{dy}}{\text{dx}}=\cos(2\text{x})$ then y =

$\frac{\sin(2\text{x)}}{\text{2}}+\text{c}$
$2\sin(2\text{x})+\text{c}$
$\frac{\sin(\text{x})}{2}+\text{c}$
None of these

Answer

$\frac{\sin(2\text{x)}}{\text{2}}+\text{c}$

Solution:
Using substitution method.
u = 2x
du = 2dx
$\frac{\text{du}}{2}=\text{dx}$
Plug in.
$\text{dy}=\frac{1}{2}\cos(\text{u})\text{du}$
Integrate.
$\text{y}=\frac{\sin(2\text{x})}{2}+\text{c}$

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Question 1161 Mark

$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\text{x}}\text{ dx}$ is equal to:

$0$
$1$
$\frac{\pi}{2}$
$\frac{\pi}{4}$

Answer

$\frac{\pi}{4}$

Solution:
We have,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\big(\frac{\pi}{2}-{\text{x}}\big)}\text{ dx}$
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\tan^3\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{1}{1+\cot^3\text{x}}+\frac{1}{\tan^3\text{x}}\Big]\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{2+\tan^3\text{x}+1+\cot^3\text{x}}{(1+\cot^3\text{x})(1+\tan^3\text{x})}\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{2\tan^3\text{x}+\cot^3\text{x}}{1+\tan^3\text{x}+\cot^3\text{x}+\cot^3\text{x}\tan^3\text{x}}\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{2\tan^3\text{x}+\cot^3\text{x}}{1+\tan^3\text{x}+\cot^3\text{x}+1}\Big]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{2+\tan^3\text{x}+\cot^3\text{x}}{2+\tan^3\text{x}+\cot^3\text{x}}\Big]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\Big[1\Big]^{\frac{\pi}{2}}_0$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$

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MCQ 1171 Mark

If $\text{I}_{10}=\int\limits^\frac{\pi}{2}_0\text{x}^{10}\sin\text{x}\text{ dx},$ then the value of $l_{10} + 90l_8$ is:

A
$9\Big(\frac{\pi}{2}\Big)^9$
✓
$10\Big(\frac{\pi}{2}\Big)^9$
C
$\Big(\frac{\pi}{2}\Big)^9$
D
$9\Big(\frac{\pi}{2}\Big)^8$

Answer

Correct option: B.

$10\Big(\frac{\pi}{2}\Big)^9$

We have,
$\text{I}_{10}=\int\limits^\frac{\pi}{2}_0\text{x}^{10}\sin\text{x}\text{ dx}$
$=\big[\text{x}^{10}(-\cos\text{x})\big]^\frac{\pi}{2}_0-\int\limits^\frac{\pi}{2}_0\big[10\text{x}^9\int\sin\text{x}\text{ dx}\big]\text{dx}$
$=\big[-\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0-10\int\limits^\frac{\pi}{2}_0\text{x}^9(-\cos\text{x})\text{dx}$
$=-\big[\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0+10\int\limits^\frac{\pi}{2}_0\text{x}^9\cos\text{x}\text{ dx}$
$=-\big[\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0+10\big[\text{x}^9\sin\text{x}\big]^\frac{\pi}{2}_0-10\int\limits^\frac{\pi}{2}_09\text{x}^8\sin\text{x}\text{ dx}$
$=-\Big[\Big(\frac{\pi}{2}\Big)^{10}\times0-0^{10}\cos0\Big]+10\Big[\Big(\frac{\pi}{2}\Big)^9\times1-0^9\times0\Big]\\-90\int\limits^\frac{\pi}{2}_0\text{x}^8\sin\text{x dx}$
$=10\Big[\Big(\frac{\pi}{2}\Big)^9\times1\Big]-90\text{I}_8$
$=10\Big(\frac{\pi}{2}\Big)^9-90\text{I}_8$
$\therefore\ \text{I}_{10}+90\text{I}_8$
$=10\Big(\frac{\pi}{2}\Big)^9$

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Question 1181 Mark

The value of $\int\limits^\pi_{-\pi}\sin^3\text{x}\cos^2\text{x}\text{ dx}$ is:

$\frac{\pi^4}{2}$
$\frac{\pi^4}{4}$
$0$
none of these

Answer

$0$

Solution:
$\int\limits^\pi_{-\pi}\sin^3\text{x}\cos^2\text{x}\text{ dx}$
$=\int\limits^\pi_{-\pi}\sin\text{x}(1-\cos^2\text{x})\cos^2\text{x}\text{ dx}$
Let $\cos\text{x}=\text{t},$ then $-\sin\text{x}\text{ dx}=\text{dt}$
When, $\text{x}=-\pi,\text{t}-1,\text{x}=\pi,\text{t}=-1$
Therefore the integral becomes
$\int\limits^{-1}_{-1}(1-\text{t}^2)\text{t}^2\text{ dt}$
$=0$

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Question 1191 Mark

$\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}$ is equal to:

$2\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
$0$
$\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$
$\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{2\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$

Answer

$\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$

Solution:
$\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$
According to the additivity property of integrals,
$\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{c}_\text{a}\text{f}(\text{x})+\int\limits^\text{b}_\text{c}\text{f}(\text{x})\text{dx},$ where a < c < b
Using this property,
$\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}\ ....(\text{i})$
Now, consider the integral, $\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}$
Let x = 2a - t Then dx = d (2a - t), dx = - dt
Also, x = a, t = a and x = 2a, t = 0
Therefore, $\int\limits^{2\text{a}}_\text{a}\text{f}(\text{x})\text{dx}=-\int\limits^0_\text{a}\text{f}(2\text{a}-\text{t})\text{dt}$
$\Rightarrow \int\limits^{2\text{a}}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{t})\text{dt}$
$\Rightarrow \int\limits^{2\text{a}}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$
Substituting this in equation (i) we get,
$\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$

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Question 1201 Mark

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

$\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$
$\frac{-\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$
$\frac{\text{e}^\text{x}}{(1+\text{e}^2)^2}+\text{C}$
$\frac{-\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{C}$

Answer

$\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$

Solution:
$\int\text{e}^\text{x}\Big(\frac{1-\text{x}}{1+\text{x}^2}\Big)^2\text{dx}$
$=\int\text{e}^\text{x}\frac{1+\text{x}^2-2\text{x}}{(1+\text{x}^2)^2}\text{dx}$
$=\int\text{e}^\text{x}\Big[\frac{1}{(1+\text{x}^2)}-\frac{2\text{x}}{(1+\text{x}^2)^2}\Big]\text{dx}$
$=\int\text{e}^\text{x}[\text{f(x)}+\text{f}'(\text{x})]\text{dx},$ where $\text{f(x)}=\frac{1}{1+\text{x}^2}$
$=\text{e}^\text{x}\text{f(x)}+\text{C}=\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$

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Question 1211 Mark

Choose the correct answer in Exercise:
The value of $\int^{1}_{0}\tan^{-1}\bigg(\frac{2\text{x}-1}{1+\text{x}-\text{x}^{2}}\bigg)\text{dx}$ is

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

Answer

0

$\text{Let I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{2\text{x}-1}{1+\text{x}-\text{x}^{2}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{\text{x}-(1-\text{x})}{1+\text{x}(1-\text{x})}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}\text{x}-\tan^{-1}(1-\text{x)}\Big]\text{dx}$
$\Rightarrow\text{I}=\int^{\text{1}}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}(1-1+\text{x)}\Big]\text{dx}$
$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$
$\Rightarrow\text{I}=\int^{1}_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$
Adding (1) and (2), we obtain
$\Rightarrow2\text{I}=\int^{1}\limits_{0}\Big(\tan^{-1}\text{x)}+\tan^{-1}(1-\text{x)}-\tan^{-1}\text{x}\Big)\text{dx}$
$\Rightarrow2\text{I}=0$
$\Rightarrow\text{I}=0$

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Question 1221 Mark

Choose the correct answer in Exercise:
$\int^{\sqrt{3}}_{1}\frac{\text{dx}}{1+\text{x}^{2}}\text{equals}$

$\frac{\pi}{3}$
$\frac{2\pi}{3}$
$\frac{\pi}{6}$
$\frac{\pi}{12}$

Answer

$\frac{\pi}{12}$

$\int\frac{\text{dx}}{1+\text{x}^{2}}=\tan^{-1}\text{x}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\int\limits_{1}^{\sqrt{3}}\frac{\text{dx}}{1+\text{x}^{2}}=\text{F}(\sqrt{3})-\text{F}(1)$
$=\tan^{-1}\sqrt{3}-\tan^{-1}1$
$=\frac{\pi}{3}-\frac{\pi}{4}$
$=\frac{\pi}{12}$

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Question 1231 Mark

What is the value of $\int_{0}^{\frac{ \pi}{2}}\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan}\text{x}+\sqrt{cot}\text{x}}\text{dx}?$

$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{\pi}{8}$
$\text{None of these}$

Answer

$\frac{\pi}{4}$

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Question 1241 Mark

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

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

Answer

$\frac{\pi}{2}$

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Question 1251 Mark

The anti derivative of $\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$ equals:

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

Answer

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

Solution:
$\int\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$
$=\int\text{x}^\frac{1}{2}\text{dx}+\int\text{x}^-\frac{1}{2}\text{dx}$
we know that $\int\text{x}^\text{n}\text{dx}=\frac{\text{x}^\text{n}+1}{\text{n}+1}$
$=\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}+\frac{\text{x}^\frac{1}{2}}{\frac{1}{2}}+\text{c}$
$=\frac{2}{3}\text{x}^\frac{3}{2}+2\text{x}^\frac{1}{2}+\text{c}$

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Question 1261 Mark

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

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

Answer

1

Solution:
We have,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_01(-\cos\text{x})\text{dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\int\limits^{\frac{\pi}{2}}_0\cos\text{x dx}$
$=-\big[\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\big[\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$=-\big[0-0\big]+\big[1-0\big]$
$=1$

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Question 1271 Mark

What is the value of $\int_{-1}^{1}\sin^3\text{x}\cos^2\text{xdx}?$

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

Answer

$0$

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Question 1281 Mark

Choose the correct answer in Exercise:$\int^\frac{2}{3}_{0}\frac{\text{dx}}{4+9\text{x}^{2}}\text{equals}$

$\frac{\pi}{6}$
$\frac{\pi}{12}$
$\frac{\pi}{24}$
$\frac{\pi}{4}$

Answer

$\frac{\pi}{24}$ $\int\frac{\text{dx}}{4+9\text{x}^{2}}=\int\frac{\text{dx}}{(2)^{2}+(3\text{x})^{2}}$$\ \text{put}\ 3\text{x}=\text{t}\Rightarrow3\text{dx}=\text{dt}$
$\therefore\int\frac{\text{dx}}{(2)^{2}+(3\text{x})^{2}}=\frac{1}{3}\int\frac{\text{dt}}{(2)^{2}+\text{t}^{2}}$
$=\frac{1}{3}\bigg[\frac{1}{2}\tan^{-1}\frac{t}{2}\bigg]$
$=\frac{1}{6}\tan^{-1}\bigg(\frac{3\text{x}}{2}\bigg)$
$=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain $\int\limits_{0}^{\frac{2}{3}}\frac{\text{dx}}{4+9\text{x}^{2}}=\text{F}\bigg(\frac{2}{3}\bigg)-\text{F}(0)$ $=\frac{1}{6}\tan^{-1}\bigg(\frac{3}{2}.\frac{2}{3}\bigg)-\frac{1}{6}\tan^{-1}0$ $=\frac{1}{6}\tan^{-1}1-0$ $=\frac{1}{6}\times\frac{\pi}{4}$ $=\frac{\pi}{24}$

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Question 1291 Mark

$\int\sec^2\text{x}.\text{cosec}^2\text{xdx}=$

$\tan\text{x}-\cot\text{x+c}$
$\tan\text{x}+\cot\text{x+c}$
$-\tan\text{x}+\cot\text{x+c}$
$\sec\text{x}\tan\text{x+c}$

Answer

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

Solution:
$\int\sec^2\text{x}.\text{cosec}^2\text{xdx}$
$=\int\frac{{1}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$
$=\int\frac{{{{\cos^2\text{x.}\sin^2\text{x}}}}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$
$=\int\frac{{{{\cos^2\text{x.}}}}}{{\cos^2\text{x}\sin^2\text{x}}}+\frac{{{{\sin^2\text{x}}}}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$
$=\int(\text{cosec}^2\text{x}+\sec^2\text{x})\text{dx}$
$=-\cot\text{x}+\tan\text{x}+\text{c}$
$=\tan\text{x}-\cot\text{x+c}$

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Question 1301 Mark

If $\int\sin\text{xd}(\sec\text{x})=\text{f}(\text{x})-\text{g}(\text{x})+\text{c},$ then:

$\text{f}(\text{x})=\sec\text{x}$
$\text{f}(\text{x})=\tan\text{x}$
$\text{g}(\text{x})=2\text{x}$
$\text{g}=-\text{x}$

Answer

$\text{f}(\text{x})=\tan\text{x}$

Solution:
$\int\sin \text{xd}(\sec\text{x})=\int\sin\text{x}\sec\text{x}\tan\text{xdx}$
$=\int\tan^2\text{xdx}$
$=\int(\sec^2\text{x - 1})\text{dx}$
$=\tan\text{x - x}+\text{c}$
$\Rightarrow\text{f}(\text{x})=\tan\text{x},\text{g}(\text{x})=\text{x}$

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Question 1311 Mark

$\int\limits_{2}^{2} \mid\text{x}\mid\text{dx}=$

0
2
1
4

Answer

4

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Question 1321 Mark

Choose the correct answer in Exercises:
$\int\frac{\text{dx}}{\sin^2\text{x}\cos^2\text{x}}\text{ equals}$

$\tan\text{x}+\cot\text{x}+\text{C}$
$\tan\text{x}-\cot\text{x}+\text{C}$
$\tan\text{x}\cot\text{x}+\text{C}$
$\tan\text{x}-\cot\text{2x}+\text{C}$

Answer

$\int\frac{\text{dx}}{\sin^2\text{x}\cos^2\text{x}}= \int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{\sin^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}+ \frac{\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{1}{\cos^2\text{x}}+\frac{1}{\sin^2\text{x}}\text{ dx}=\int\sec^2\text{x dx}+\int\text{cosec}^2\text{x}\text{ dx}$
$=\int\sec^2\text{x}\text{ dx} +\text{ dx}\int\ \text{cosec}^2\text{x}\text{ dx}$
$=\tan\text{x}-\cot\text{x}+\text{c} $
Therefore, option (B) is correct.

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Question 1331 Mark

Choose the correct answer in Exercises:
$\int\frac{10\text{x}^9+10^{\text{x}}\log_\text{e}10}{\text{x}^{10}+10^{\text{x}}}\text{ equals}$

$10^\text{x}-\text{x}^{10}+\text{C}$
$10^\text{x}+\text{x}^{10}+\text{C}$
$(10^\text{x}-\text{x}^{10})^{-1}+\text{C}$
$\log(10^\text{x}+\text{x}^{10})+\text{C}$

Answer

$\text{Let I}=\int\frac{10\text{x}^9+10^{\text{x}}\log_\text{e}10}{\text{x}^{10}+10^{\text{x}}}\text{ dx} \ \ \ \ ...\text{(i)} $
Putting ${\text{x}^{10}+10^{\text{x}}}=\text{t}\ \ \ \ \Rightarrow\ \ \ \ (10\text{x}^9+10^{\text{x}}\log_\text{e}10)\text{ dx = dt} $
$\therefore\ \ \ \ \ $From eq. (i), $\text{I}=\int\frac{\text{dt}}{\text{t}}=\log\begin{vmatrix}\text {t}\end{vmatrix}+\text{c}=\log\begin{vmatrix}\text{x}^{10}+10^\text{x}\end{vmatrix}+\text{c}$
Therefore, option (D) is correct.

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Question 1341 Mark

Choose the correct answer in Exercise:
$\int\frac{\text{dx}}{\sqrt{9\text{x}-4\text{x}^2}}\text{equals}$

$\frac{1}{9}\sin^{-1}\bigg(\frac{9\text{x}-8}{8}\bigg)+\text{C}$
$\frac{1}{2}\sin^{-1}\bigg(\frac{8\text{x}-9}{9}\bigg)+\text{C}$
$\frac{1}{3}\sin^{-1}\bigg(\frac{9\text{x}-8}{8}\bigg)+\text{C}$
$\frac{1}{2}\sin^{-1}\bigg(\frac{9\text{x}-8}{9}\bigg)+\text{C}$

Answer

$\text{Let I}=\int\frac{\text{dx}}{\sqrt{9\text{x}-4\text{x}^2}}$
$=\int\frac{1}{\sqrt{-4\text{x}^2+9\text{x}}}\text{ dx}$
$=\int\frac{1}{-4\bigg(\text{x}^2-\frac{9}{4}\text{x}\bigg)}\text{ dx}$
$=\int\frac{1}{-4\Bigg[\text{x}^2-\frac{9}{4}\text{x}+\bigg(\frac{9}{8}\bigg)^2-\bigg(\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\int\frac{1}{-4\Bigg[\bigg(\text{x}-\frac{9}{8}\bigg)^2+\bigg(\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\int\frac{1}{4\Bigg[\bigg(\frac{9}{8}\bigg)^2-\bigg(\text{x}-\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\frac{1}{2}\int\frac{1}{\Bigg[\bigg(\frac{9}{8}\bigg)^2-\bigg(\text{x}-\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\frac{1}{2}\sin^{-1}\frac{\text{x}-\frac{9}{8}}{\frac{9}{8}}+\text{C}$
$=\frac{1}{2}\sin^{-1}\bigg(\frac{8\text{x}-9}{9}\bigg)+\text{C}$
Therefore, option (B) is correct.

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Question 1351 Mark

Choose the correct answer in Exercise:
$\int\frac{\text{dx}}{\text{x}^2+2\text{x}+2}\text{equals}$

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

Answer

$\int\frac{\text{dx}}{\text{x}^2+2\text{x}+2}$
$=\int\frac{1}{\text{x}^2+2\text{x}+1+1}\text{dx}$
$=\int\frac{1}{(\text{x+1})^2+(1)^2}\text{ dx}$
$=\tan^{-1}(\text{x}+1)+\text{C}$
Therefore, option (B) is correct.

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Question 1361 Mark

Choose the correct answer in Exercises:
$\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^2(\text{e}^\text{x}\text{x})}\text{dx}$ equals

$-\cot(\text{e}^{\text{x}}\text{x})+\text{C}$
$\tan(\text{xe}^\text{x})+\text{C}$
$\tan(\text{e}^\text{x})+\text{C}$
$\cot(\text{e}^\text{x})+\text{C}$

Answer

$\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^2(\text{e}^\text{x}\text{x})}\text{dx}$
$\text{Let }\text{e}^{\text{x}}\text{x}=\text{t}$
$\Rightarrow(\text{e}^\text{x}\cdot\text{x}+\text{e}^{\text{x}}\cdot1)\text{dx}=\text{dt}$
$\text{e}^{\text{x}}(\text{x}+1)\text{dx}=\text{dt}$
$\therefore\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^2(\text{e}^\text{x}\text{x})}\text{dx}=\int\frac{\text{dt}}{\cos^2\text{t}}$
$=\int\sec^2\text{t}\text{ dt}$
$=\tan\text{t}+\text{C}$
$=\tan(\text{e}^{\text{x}}\cdot\text{x})+\text{C}$
Hence, the correct answer is B.

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Question 1371 Mark

Choose the correct answer in Exercises:
The anti derivative of $\bigg(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\bigg)$equals

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

Answer

$\int\bigg(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\bigg)\text{ dx}=\int\bigg(\text{x}^{\frac{1}{2}}+\text{x}^{\frac{-1}{2}}\bigg)\text{ dx} $ $=\int\text{x}^\frac{1}{2}\text{ dx}+\int\text{x}^\frac{-1}{2}\text{ dx}=\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1} +\frac{\text{x}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\text{C} $ $=\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}} +\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{C}$$=\frac{2}{3}\text{x}^\frac{3}{2}+2\text{x}^\frac{1}{2}+\text{C}$

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Question 1381 Mark

Choose the correct answer in Exercises:
$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$ is equal to

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

Answer

$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\bigg(\frac{\sin^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}-\frac{\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int(\sec^2\text{x}-\text{cosec}^2\text{x})\text{ dx}$
$=\tan\text{x}+\cot\text{x}+\text{C}$
Hence, the correct answer is A.

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Question 1391 Mark

Choose the correct answer in Exercises:If $\frac{\text{d}}{\text{dx}}\text{f}\text{(x)}=4\text{x}^3-\frac{3}{\text{x}^4}$such that $\text{f}(2)=0.$Then $\text{f}\text{(x)}$ is

$\text{x}^4+\frac{1}{\text{x}^3}-\frac{129}{8}$
$\text{x}^3+\frac{1}{\text{x}^4}+\frac{129}{8}$
$\text{x}^4+\frac{1}{\text{x}^3}+\frac{129}{8}$
$\text{x}^3+\frac{1}{\text{x}^4}-\frac{129}{8}$

Answer

$\text{f(x)}=\int\bigg(4\text{x}^3-\frac{3}{\text{x}^4}\bigg)\text{ dx}$
$=4\int\text{x}^3\text{ dx}-3\int\frac{1}{\text{x}^4}\text{ dx} $
$=4.\frac{\text{x}^4}{4}-3\int\text{x}^{-4}\text{ dx} =\text{x}^4-3\frac{\text{x}^-3}{-3}+\text{c} $
$\Rightarrow\ \ \ \ \ \ \ \text{f(x)}=\text{x}^4+\frac{1}{\text{x}^3}+\text{c} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ....\text{(i)} $
$\Rightarrow\ \ \ \ \ \ \ \text{f(2)}=16+\frac{1}{8}+\text{c} \ \ \ \ \Rightarrow\ \ \ \ \ \ 0=\frac{128+1}{8}+\text{c} $
$\Rightarrow\ \ \ \ \ \ \ \text{c }+\frac{129}{8}=0 \ \ \ \ \Rightarrow\ \ \ \ \ \ \text{c }=\frac{-129}{8} $
Putting $\text{c}=\frac{-129}{8}$ in eq. (i),
$\text{f(x)}=\text{x}^4+\frac{1}{\text{x}^3}-\frac{129}{8} $
Therefore, option (A) is correct.

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Question 1401 Mark

Evaluate: $\int(2\tan\text{x}-3\cot\text{x})^2\text{dx}.$

$-4\tan\text{x}-\cot\text{x}-25\text{x}+\text{c}$
$4\tan\text{x}-9\cot\text{x}-25\text{x}+\text{c}$
$-4\tan\text{x}+9\cot\text{x}+25\text{x}+\text{c}$
$4\tan\text{x}+9\cot\text{x}+25\text{x}+\text{c}$

Answer

$4\tan\text{x}+9\cot\text{x}+25\text{x}+\text{c}$

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Question 1411 Mark

If $\int\text{f}(\text{x})\text{dx}=-2\cos\sqrt{\text{x}}+\text{c}$ then f(x) is equal to:

$\sin\sqrt{\text{x}}$
$\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}$
$2\cos\sqrt{\text{x}}$
None of these

Answer

$\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}$

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Question 1421 Mark

$\int\frac{\cos2\text{x dx}}{(\sin\text{x}+\cos\text{x})^2}=$

$-\frac{1}{\sin\text{x}+\cos\text{x}}+\text{c}$
$\log|\sin\text{x}+\cos\text{x }|+\text{c}$
$\log|\sin\text{x}-\cos\text{x }|+\text{c}$
$\frac{1}{(\sin\text{x}+\cos\text{x})^2}$

Answer

$\log|\sin\text{x}+\cos\text{x }|+\text{c}$

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Question 1431 Mark

If $\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}\cos\text{xdx}$ is equal to:

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

Answer

$\frac{1}{2}$

Solution:
$\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}\cos\text{xdx}$
$\sin\text{x}=\text{t}\Rightarrow\cos\text{xdx}=\text{dt}$
$\text{x}\Rightarrow0\Rightarrow\frac{\pi}{2}$
$\int\limits_{0}^{\frac{\pi}{2}}\text{tdt}$
$\Rightarrow\frac{\text{t}^2}{2}\mid^1_0$
$\Rightarrow\frac{1}{2}-0=\frac{1}{2}$

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Question 1441 Mark

The value of $\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\big(\text{x}^3+\text{x}\cos\text{x}+\tan^5\text{x}+1\big)\text{dx},$ is:

0
2
$\pi$
1

Answer

$\pi$

Solution:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\big(\text{x}^3+\text{x}\cos\text{x}+\tan^5\text{x}+1\big)\text{dx}$
$=\Big[\frac{\text{x}^4}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}+\Big[\text{x}\sin\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin\text{x}\text{dx}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}(\sec^2\text{x}-1)\text{dx}+\Big[\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\frac{\pi^4}{64}-\frac{\pi^4}{64}+\frac{\pi}{2}-\frac{\pi}{2}-\Big[-\cos\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\text{ dx}+\frac{\pi}{2}+\frac{\pi}{2}$
$=+0+\Big[\frac{\tan^4\text{x}}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\text{ dx}$
$=\pi-\Big[\frac{\tan^2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\Big[-\log\big(\cos\text{x}\big)\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\pi$

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Question 1451 Mark

The value of $\int\frac{\cos2\text{x}}{{\cos}{\text{ x}}}\text{dx}$ is equal to:

$2\sin\text{x}-\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$
$2\sin\text{x}-\ell\text{ n }\mid\sec\text{x}-\tan\text{x}\mid+\text{ c}$
$2\sin\text{x}+\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$
$3\sin\text{x}-\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$

Answer

$2\sin\text{x}-\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$

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Question 1461 Mark

Choose the correct answer in exercise:
$\int\frac{\text{x dx}}{(\text{x}-1)(\text{x}-2)}$ equals

$\text{log}\Bigg|\frac{(\text{x}-1)^2}{\text{x}-2}\Bigg|+\text{C}$
$\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{C}$
$\text{log}\Bigg|\Bigg(\frac{\text{x}-1}{\text{x}-2}\Bigg)^2\Bigg|+\text{C}$
$\text{log}|(\text{x}-1)(\text{x}-2)|+\text{c}$

Answer

$\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{C}$

Let, $\frac{\text{x}}{(\text{x}-1)(\text{x}-2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-2}\dots(\text{i})$
⇒ x = A(x - 2) + B(x -1)
⇒ x = Ax - 2A + Bx - B
Comparing coefficients of x:
A + B = 1 ……….(ii)
Comparing constants
–2A – B = 0 ……….(iii)
On solving eq. (ii) and (iii), we get
A = –1, B = 2
Putting these values of A and B in eq. (i),
$\frac{\text{x}}{(\text{x}-1)(\text{x}-2)}=\frac{-1}{\text{x}-1}+\frac{2}{\text{x}-2}$
$\Rightarrow\ \int\frac{\text{x}}{(\text{x}-1)(\text{x}-2)}\text{dx}=-\int\frac{1}{\text{x}-1}\text{dx}+2\int\frac{1}{\text{x}-2}\text{dx}$
$=-\text{log}|\text{x}-1|+2\text{log}|\text{x}-2|+\text{c}$
$=\text{log}|(\text{x}-2)^2|-\text{log}|\text{x}-1|+\text{c}=\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{c}$

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Question 1471 Mark

Evaluate: $\int\frac{1-\cos\text{x}}{cos\text{x}(1+cos\text{x})}\text{dx}.$

$\log|\text{sec}\text{x}+\tan\text{x}|-2\tan(\frac{\text{x}}{2})+\text{c}$
$\log|\text{sec}\text{x}-\tan\text{x}|-2\tan(\frac{\text{x}}{2})+\text{c}$
$\log|\text{sec}\text{x}+\tan\text{x}|+2\tan(\frac{\text{x}}{2})+\text{c}$
$\text{None of these}$

Answer

$\log|\text{sec}\text{x}+\tan\text{x}|-2\tan(\frac{\text{x}}{2})+\text{c}$

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Question 1481 Mark

$\int\limits^1_0\frac{\text{x}}{(1-\text{x})^{54}}\text{ dx}=$

$\frac{15}{16}$
$\frac{3}{16}$
$-\frac{3}{16}$
$-\frac{16}{3}$

Answer

$-\frac{16}{3}$

solution:
$\text{I}=\int\limits^1_0\frac{\text{x}}{(1-\text{x})^{\frac{5}{4}}}\text{ dx}$
Put, 1 - x = t ⇒ x =1 -t
⇒ dx = -dt

x	0	1
t	1	0

$\text{I}=\int\limits^0_1\frac{(1-\text{t})(-\text{dt})}{\text{t}^{\frac{5}{4}}}$
$\text{I}=\int\limits^1_0\Big(\text{t}^\frac{5}{4}-\text{t}^\frac{-1}{4}\Big)\text{dt}$
$\text{I}=\Bigg[\frac{\text{t}^{-\frac{5}{4}}}{\frac{-1}{4}}-\frac{\text{t}^\frac{3}{4}}{\frac{3}{4}}\Bigg]^1_0$
$\text{I}=-4-\frac{4}{3}$
$\text{I}=\frac{-16}{3}$

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Question 1491 Mark

Choose the correct option from given four options:
$\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\sin2\text{x}}\text{ dx}$ is equal to:

$2\sqrt{2}$
$\big(\sqrt{2}+1)$
$2$
$2\big(\sqrt{2}-1)$

Answer

$2\big(\sqrt{2}-1)$

Solution:
Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\sin2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\sqrt{(\cos\text{x}-\sin\text{x})^2}\text{ dx}+\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}}\sqrt{(\sin\text{x}-\cos\text{x})^2}\text{ dx}$
$=\Big[\sin\text{x}+\cos\text{x}\Big]^{\frac{\pi}{4}}_0+\Big[-\cos\text{x}-\sin\text{x}\Big]^{\frac{\pi}{2}}_{\frac{\pi}{4}}$
$=\frac{1}{\sqrt{2}}+=\frac{1}{\sqrt{2}}-0-1+\Big(-0-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\Big)$
$=2\sqrt{2}-2=2(\sqrt{2}-1)$

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Question 1501 Mark

If $\int\frac{1}{5+4\sin\text{x}}\text{ dx}=\text{A}\tan^{-1}\Big(\text{B}\tan\frac{\pi}{2}+\frac{4}{3}\Big)+\text{C},$ then:

$\text{A}=\frac{2}{3},\text{B}=\frac{5}{3}$
$\text{A}=\frac{1}{3},\text{B}=\frac{2}{3}$
$\text{A}=-\frac{2}{3},\text{B}=\frac{5}{3}$
$\text{A}=\frac{1}{3},\text{B}=-\frac{5}{3}$

Answer

$\text{A}=\frac{2}{3},\text{ B}=\frac{5}{3}$

Solution:
$\int\frac{1}{5+4\sin\text{x}}\text{ dx}=\text{A}\tan^{-1}\Big(\text{B}\tan\frac{\pi}{2}+\frac{4}{3}\Big)+\text{C}\ ...(\text{i})$
Considering the LHS of eq. (1)
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\Rightarrow\int\frac{1}{5+\frac{8\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{dx}$
$\Rightarrow\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+8\tan\frac{\text{x}}{2}}\text{dx}$
$\Rightarrow\frac{\sec^2\frac{\text{x}}{2}}{5\tan^2\frac{\text{x}}{2}+8\tan\frac{\text{x}}{2}+5}\text{dx}\dots(2)$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\Rightarrow\sec^2\frac{\text{x}}{2}\times\frac{1}{2}\text{dx}=\text{dt}$
$\Rightarrow\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{2dt}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{8}{5}\text{t}+1}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{8}{5}\text{t}+\big(\frac{4}{5}\big)^2-\big(\frac{4}{5}\big)^2+1}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\big(\text{t}+\frac{4}{5}\big)^2+1-\frac{16}{25}}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\big(\text{t}+\frac{4}{5}\big)^2+\big(\frac{3}{5}\big)^2}$
$\Rightarrow\frac{2}{5}\times\frac{2}{3}\tan^{-1}\bigg(\frac{\text{t}+\frac{4}{5}}{\frac{3}{5}}\bigg)+\text{C}$
$\Rightarrow\frac{2}{3}\tan^{-1}\Big(\frac{5\text{t}+4}{3}\Big)+\text{C}$
$\Rightarrow\frac{2}{3}\tan^{-1}\Big(\frac{5}{3}\tan\frac{\text{x}}{2}+\frac{4}{3}\Big)+\text{C}\ ...\text{(ii)}$ $\Big(\because\text{t}=\tan\frac{\text{x}}{2}\Big)$
Comparing eq. (ii) with the RHS of eq. (i) we get,
$\therefore\ \text{A}=\frac{2}{3},\text{ B}=\frac{5}{3}$

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M.C.Q (1 Marks) - Page 3 - MATHS STD 12 Science Questions - Vidyadip