Solve the Following Question.(4 Marks)

Question 14 Marks

Evaluate the following:

$\int_0^{\pi / 4} \frac{\cos 2 x}{1+\cos 2 x+\sin 2 x} d x$

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Question 24 Marks

Evaluate the following:

$\int_0^\pi \frac{x}{1+\sin ^2 x} d x$

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Question 34 Marks

Evaluate the following:

$\int_0^\pi x \cdot \sin x \cdot \cos ^4 x d x$

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Question 44 Marks

Evaluate the following:
$\int_{-1}^1 \frac{1+x^3}{9-x^2} d x$

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Question 54 Marks

Evaluate the following:

$\int_{\pi / 4}^{\pi / 2} \frac{\cos \theta}{\left[\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right]^3} d \theta$

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Question 64 Marks

Evaluate the integral $\int_0^\pi \cos ^2 x \cdot d x$ using the result/ property.

Answer

$
\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-x) d x
$
Let,
$
\begin{aligned}
& \mathrm{I}=\int_0^\pi \cos ^2 x \cdot d x \\
& =\int_0^{2\left(\frac{\pi}{2}\right)} \cos ^2 x \cdot d x \\
& =\int_0^{\pi / 2} \cos ^2 x \cdot d x+\int_0^{\pi / 2}\left[\cos \left(2 \frac{\pi}{2}-x\right)\right]^2 \cdot d x \\
& =\int_0^{\pi / 2} \cos ^2 x \cdot d x+\int_0^{\pi / 2} \cos ^2 x \cdot d x \\
& \because \cos (\pi-x)=-\cos x \\
& =2 \cdot \int_0^{\pi / 2} \cos ^2 x \cdot d x \\
& =\int_0^{\pi / 2}(1+\cos 2 x) \cdot d x \\
& =\left[x+\sin 2 x \cdot \frac{1}{2}\right]_0^{\pi / 2} \\
& =\left[\left(\frac{\pi}{2}+\frac{1}{2} \sin 2 \frac{\pi}{2}\right)-\left(0+\frac{1}{2} \sin 2(0)\right)\right] \\
& =\frac{\pi}{2}+0 \quad \because \sin 0=0 ; \sin \pi=0 \\
& =\frac{\pi}{2} \\
& \therefore \int_0^\pi \cos ^2 x \cdot d x=\frac{\pi}{2} \\
\end{aligned}
$

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Question 74 Marks

Evaluate : $\int_0^{\pi / 2} \frac{1}{1+\sqrt[3]{\tan x}} \cdot d x$

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Question 84 Marks

Evaluate : $\int_{-1}^1|5 x-3| \cdot d x$

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Question 94 Marks

Evaluate : $\int_0^{1 / 2} \frac{1}{\left(1-2 x^2\right) \sqrt{1-x^2}} \cdot d x$

Answer

Let $\quad I=\int_0^{1 / 2} \frac{1}{\left(1-2 x^2\right) \sqrt{1-x^2}} \cdot d x$
put $x=\sin \theta \quad \therefore \quad 1 \cdot d x=\cos \theta \cdot d \theta$
As $\quad x$ varies from 0 to $\frac{1}{2}, \theta$ varies from 0 to $\frac{\pi}{6}$
$
\begin{aligned}
& =\int_0^{\pi / 6} \frac{\cos \theta}{\left(1-2 \sin ^2 \theta\right) \sqrt{1-\sin ^2 \theta}} \cdot d \theta=\int_0^{\pi / 6} \frac{\cos \theta}{(\cos 2 \theta) \sqrt{\cos ^2 \theta}} \cdot d \theta \\
& =\int_0^{\pi / 6} \frac{1}{\cos 2 \theta} \cdot d \theta \\
& =\int_0^{\pi / 6} \sec 2 \theta \cdot d \theta \\
& =\left[\log (\sec 2 \theta+\tan 2 \theta) \cdot \frac{1}{2}\right]_0^{\pi / 6} \\
& =\frac{1}{2} \cdot\left[\log \left(\sec 2\left(\frac{\pi}{6}\right)+\tan 2\left(\frac{\pi}{6}\right)-\log (\sec 0+\tan 0)\right]\right. \\
& =\frac{1}{2} \cdot\left[\log \left(\sec \frac{\pi}{3}+\tan \frac{\pi}{3}\right)-\log (1+0)\right] \\
& =\frac{1}{2} \cdot[\log (2+\sqrt{3})-0] \\
& =\frac{1}{2} \log (2+\sqrt{3})
\end{aligned}
$
$
\therefore \quad \int_0^{1 / 2} \frac{1}{\left(1-2 x^2\right) \sqrt{1-x^2}} \cdot d x=\frac{1}{2} \log (2+\sqrt{3})
$

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Question 104 Marks

Evaluate : $\int_0^{\pi / 2} \frac{\cos x}{1+\cos x+\sin x} \cdot d x$

Answer

$
\begin{aligned}
\text { Let I } & =\int_0^{\pi / 2} \frac{\cos x}{1+\cos x+\sin x} \cdot d x \\
& =\int_0^{\pi / 2} \frac{\cos ^2\left(\frac{x}{2}\right)-\sin ^2\left(\frac{x}{2}\right)}{2 \cos ^2\left(\frac{x}{2}\right)+2 \sin \left(\frac{x}{2}\right) \cdot \cos \left(\frac{x}{2}\right)} \cdot d x \\
& =\int_0^{\pi / 2} \frac{\left[\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]}{2\left[\cos \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]} \cdot d x \\
& =\int_0^{\pi / 2}\left[\frac{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}\right] \cdot d x=\int_0^{\pi / 2}\left[1-\tan \left(\frac{x}{2}\right)\right] \cdot d x
\end{aligned}
$
$\begin{aligned} & =\frac{1}{2} \cdot\left[x-\log \left(\sec \frac{x}{2}\right) \cdot \frac{1}{\frac{1}{2}}\right]_0^{\pi / 2} \\ & =\frac{1}{2} \cdot\left[\frac{\pi}{2}-2 \cdot \log \left(\sec \frac{\pi}{4}\right)-(0-2 \log \sec 0)\right] \\ & =\frac{1}{2} \cdot\left[\frac{\pi}{2}-2 \log \sqrt{2}-0+2(0)\right] \quad=\frac{1}{2} \cdot\left[\frac{\pi}{2}-2 \log \sqrt{2}\right] \quad=\frac{\pi}{4}-\log \sqrt{2} \\ \therefore \quad & \int_0^{\pi / 2} \frac{\sec ^2 x}{1+\cos x+\sin x} \cdot d x=\frac{\pi}{4}-\log \sqrt{2}\end{aligned}$

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Question 114 Marks

Evaluate : $\int_0^{\pi / 4} \frac{\sec ^2 x}{2 \tan ^2 x+5 \tan x+1} \cdot d x$

Answer

Let $\mathrm{I}=\int_0^{\pi / 4} \frac{\sec ^2 x}{2 \tan ^2 x+5 \tan x+1} \cdot d x$
put $\tan x=t \quad \therefore \sec ^2 x \cdot d x=1 \cdot d t$
As $x$ varies from 0 to $\frac{\pi}{4}$
$t$ varies from 0 to 1
$
\begin{aligned}
& =\int_0^1 \frac{1}{2 t^2+4 t+1} \cdot d t \\
& =\frac{1}{2} \cdot \int_0^1 \frac{1}{t^2+2 t+\frac{1}{2}} \cdot d t \\
& =\frac{1}{2} \cdot \int_0^1 \frac{1}{t^2+2 t+1-1+\frac{1}{2}} \cdot d t \\
& =\frac{1}{2} \cdot \int_0^1 \frac{1}{(t+1)^2-\left(\frac{1}{\sqrt{2}}\right)^2} \cdot d t
\end{aligned}
$
$\begin{aligned} & =\frac{1}{2} \frac{1}{2\left(\frac{1}{\sqrt{2}}\right)}\left[\log \left[\frac{(t+1)-\frac{1}{\sqrt{2}}}{(t+1)+\frac{1}{\sqrt{2}}}\right]\right]_0^1 \\ & =\frac{\sqrt{2}}{4} \log \left[\left(\frac{\sqrt{2} t+\sqrt{2}-1}{\sqrt{2} t+\sqrt{2}+1}\right)\right]_0^1 \\ & =\frac{\sqrt{2}}{4}\left[\log \left(\frac{\sqrt{2}(1)+\sqrt{2}-1}{\sqrt{2}(1)+\sqrt{2}+1}\right)-\log \left(\frac{\sqrt{2}(0)+\sqrt{2}-1}{\sqrt{2}(0)+\sqrt{2}+1}\right)\right] \\ & =\frac{\sqrt{2}}{4}\left[\log \left(\frac{2 \sqrt{2}-1}{2 \sqrt{2}+1}\right)-\log \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\right] \\ & =\frac{\sqrt{2}}{4} \log \left[\left(\frac{2 \sqrt{2}-1}{2 \sqrt{2}+1}\right) \div\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\right] \\ & =\frac{\sqrt{2}}{4} \log \left[\frac{3+\sqrt{2}}{3-\sqrt{2}}\right]\end{aligned}$

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Question 124 Marks

Evalaute : $\int_2^3 7^x \cdot d x$

Answer

$
\begin{aligned}
& \text { Given, } \int_2^3 7^x \cdot d x=\int_a^b f(x) d x \\
& f(x)=7^x \quad a=2 ; b=3 \\
& \Rightarrow \quad f(a+r h)=f(1+r h) \\
& =7^{2+r h} \\
& =7^2 \cdot 7^{\text {th }} \\
& h=\frac{b-a}{n} \\
& h=\frac{3-2}{n} \\
& \therefore \quad n h=1 \\
\end{aligned}
$
$
\therefore \quad n h=1
$
We know $\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot f(a+r h)$
$
\therefore \quad \begin{aligned}
\int_1^3 7^x \cdot d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot\left(7^2 \cdot 7^{r h}\right) \\
& =\lim _{n \rightarrow \infty} 7^2 \cdot \sum_{r=1}^n h \cdot 7^{r \cdot h} \\
& =\lim _{n \rightarrow \infty} 7^2 \cdot h \cdot\left[7^h+7^{2 h}+7^{3 h}+7^{4 h}+\ldots+7^{h h}\right] \\
& =\lim _{n \rightarrow \infty} 7^2 \cdot h \cdot\left(\frac{7^h\left[\left(7^h\right)^n-1\right]}{7^h-1}\right)=\lim _{n \rightarrow \infty} 7^2 \cdot\left(\frac{7^h\left(7^{n h}-1\right)}{\left.\frac{7^h-1}{h}\right)}\right. \\
& =\lim _{n \rightarrow \infty} 7^2 \cdot\left(\frac{7^h\left(7^{(1)}-1\right)}{\left.\frac{7^h-1}{h}\right)}\right. \\
& =\frac{7^2 \cdot 7^0 \cdot(7-1)}{\log 7}=\frac{(49)(1)(6)}{\log 7}=\frac{294}{\log 7}
\end{aligned}
$

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Question 134 Marks

Evaluate :

$\int_0^\pi \sin ^3 x(1+2 \cos x)(1+\cos x)^2 \cdot d x$

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Question 144 Marks

Evaluate :

$\int_0^1 \frac{\log (x+1)}{x^2+1} \cdot d x$

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Question 154 Marks

Evaluate :

$\int_0^a \frac{1}{x+\sqrt{a^2-x^2}} \cdot d x$