3 Marks Question

Question 3013 Marks

Evaluate the following integrals:
$\int2\text{x}\sec^3\big(\text{x}^2+3\big)\tan\big(\text{x}^2+3\big)\text{dx}$

Answer

$\int2\text{x}\sec^3\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)\text{dx}$ $=\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$ Let $\sec\big(\text{x}^2+3\big)=\text{t}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}\text{ dx}=\text{dt}$Now, $\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{\sec^2(\text{x}^2+3)}{3}+\text{C}$

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Question 3023 Marks

Evaluate the following integrals:
$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$

Answer

$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$
$\text{Let}\ \text{log}\sin\text{x} =\text{t}$
$\Rightarrow\frac{1}{\sin\text{x}}\times \cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx} =\text{dt}$
$\text{Now},\int\cot\text{x}. \log\sin\text{x dx} $
$=\int\text{t}.\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{C}$
$=\frac{(\text{log}\begin{vmatrix}\sin\text{x}\end{vmatrix})^{2}}{2}+\text{C}$

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Question 3033 Marks

Evaluate the following integrals:
$\int\frac{1}{1-\sin\text{x}}\text{dx}$

Answer

$\int\frac{1}{1-\sin\text{x}}\text{dx}$$=\int\frac{(1+\sin\text{x})}{(1-\sin\text{x})\times(1+\sin\text{x})}\text{dx}$
$=\Big(\frac{1+\sin\text{x}}{1-\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1+\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\Big)\text{dx}$
$=\int(\sec^2\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
$=\tan\text{x}+\sec\text{x}+\text{C}$

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Question 3043 Marks

Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$

Answer

Let $\text{I}=\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\sqrt{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Bigg[\frac{2\text{t}^{\frac{3}{2}}}{3}\Bigg]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{\pi}{4}\Big)^{\frac{3}{2}}$
$\Rightarrow\text{I}=\frac{1}{12}\pi^{\frac{3}{2}}$

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Question 3053 Marks

Evalute the following integrals:
$\int\frac{1}{\text{x}(3+\log\text{x})}\text{dx}$

Answer

Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.Let $\text{I}=\int\frac{1}{\text{x}(3+\log\text{x})}\text{dx}$
Putting $\log\tan\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{dx}}{\text{x}}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{3+\text{t}}$
$=\log|3+\text{t}|+\text{C}$
$=\log|3+\log\text{x}|+\text{C}\ \big[\because\text{t}=\log\text{x}\big]$

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Question 3063 Marks

Evaluate the definite integral in Exercise:$\int_{1}^{2}(4\text{x}^{3}-5\text{x}^{2}+6\text{x}+9)\text{dx}$

Answer

$\text{Let}=\int\limits_{1}^{2}(4\text{x}^{3}-5\text{x}^{2}+6\text{x}+9)\text{dx}$$\int(4\text{x}^{3}-5\text{x}^{2}+6\text{x}+9)\text{dx}=4\bigg(\frac{\text{x}^{4}}{4}\bigg)-5\bigg(\frac{\text{x}^{3}}{3}\bigg)+6\bigg(\frac{\text{x}^{2}}{2}\bigg)+9\text{(x)}$
$=\text{x}^{4}-\frac{5\text{x}^{3}}{3}+3\text{x}^{2}+9\text{x}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(2)-\text{F}(1)$
$\text{I}=\Bigg\{2^{4}-\frac{5.(2)^{3}}{3}+3(2)^{2}+9(2)\Bigg\}-\bigg\{(1)^{4}-\frac{5(1)^{3}}{3}+3(1)^{2}+9(1)\bigg\}$
$=\bigg(16-\frac{40}{3}+12+18\bigg)-\bigg(1-\frac{5}3{+3+9}\bigg)$
$=16-\frac{40}{3}+12+18-1+\frac{5}{3}-3-9$
$=33-\frac{35}{3}$
$=\frac{99-35}{3}$
$=\frac{64}{3}$

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Question 3073 Marks

Evaluate the definite integral in Exercise:
$\int_{0}^{2}\frac{6\text{x}+3}{\text{x}^{2}+4}\text{dx}$

Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{2}\frac{6\text{x}+3}{\text{x}^{2}+4}\text{dx}$$\int\frac{6\text{x}+3}{\text{x}^{2}+4}\text{dx}=3\int\frac{2\text{x}+1}{\text{x}^{2}+4}\text{dx}$
$=3\int\frac{2\text{x}}{\text{x}^{2}+4}\text{dx}+3\int\frac{1}{\text{x}^{2}+4}\text{dx}$
$=3\log(\text{x}^{2}+4)+\frac{3}{2}\tan^{-1}\frac{\text{x}}{2}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(2)-\text{F}(0)$
$=\left\{3\log(2^{2}+4)+\frac{3}{2}\tan^{-1}\bigg(\frac{2}{2}\bigg)\right\}-\left\{3\log(0+4)+\frac{3}{2}\tan^{-1}\bigg(\frac{0}{2}\bigg)\right\}$
$=3\log8+\frac{3}{2}\tan^{-1}1-3\log4-\frac{3}{2}\tan^{-1}0$
$=3\log8+\frac{3}{2}\bigg(\frac{\pi}{4}\bigg)-3\log4-0$
$=3\log\bigg(\frac{8}{4}\bigg)+\frac{3\pi}{8}$
$=3\log2+\frac{3\pi}{8}$

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Question 3083 Marks

Evaluate the following integrals:$\int\text{x}^3\cos\text{x}^2\text{dx}$

Answer

Let $\text{I}=\int\text{x}^3\cos\text{x}^2\text{dx}$
Let $\text{x}^2=\text{t}$
$2\text{x dx = dt}$
$\text{I}=\frac{1}{2}\int\text{t}\cos\text{t dt}$
Using integration by parts,
$=\frac{1}{2}[\text{t}\int\cos\text{t dt}-\int(1\times\int\cos\text{t dt})\text{dt}]$
$=\frac{1}{2}[\text{t}\times\sin\text{t}-\int\sin\text{t dt}]$
$=\frac{1}{2}[\text{t}\sin\text{t}+\cos\text{t}]+\text{C}$
$\text{I}=\frac{1}{2}[\text{x}^2\sin\text{x}^2+\cos\text{x}^2]+\text{C}$

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Question 3093 Marks

Evaluate the following:
$\int\limits^1_0\frac{\text{x}}{\sqrt{1+\text{x}^2}}\text{dx}$

Answer

Let $\text{I}=\int\limits^1_0\frac{\text{x}}{\sqrt{1+\text{x}^2}}\text{dx}$
Put $1+\text{x}^2=\text{t}^2$
$\Rightarrow\ 2\text{xdx}=2\text{tdt}$
$\Rightarrow\ \text{xdx}=\text{tdt}$
As $\text{x}\rightarrow0,$ then $\text{t}\rightarrow1$
and $\text{x}\rightarrow\pi,$ then $\text{t}\rightarrow\sqrt{2}$
Substituting $1+\text{x}^2=\text{t}^2$ and $\text{xdx}=\text{tdt}$ in I, we get
$\therefore\ \text{I}=\int\limits^{\sqrt{2}}_0\frac{\text{tdt}}{\text{t}}$
$=[\text{t}]^{\sqrt{2}}_1=\sqrt{2}-1$

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MATHS 3 Marks Question