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Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{3+2 \cos ^{2} x} d x$

Solution:

Given I $=\int \frac{1}{3+2 \cos ^{2} x} d x$

Divide numerator and denominator by $\cos ^{2} x$,

$\Rightarrow I=\int \frac{1}{3+2 \cos ^{2} x} d x=\int \frac{\sec ^{2} x}{3 \sec ^{2} x+2} d x$

Replacing $\sec ^{2} x$ in denominator by $1+\tan ^{2} x$,

$\Rightarrow \int \frac{\sec ^{2} x}{3 \sec ^{2} x+2} d x=\int \frac{\sec ^{2} x}{3+3 \tan ^{2} x+2} d x$

$=\int \frac{\sec ^{2} x}{5+3 \tan ^{2} x} d x$

Putting $\tan x=t$ so that $\sec ^{2} x d x=d t$

$\Rightarrow \int \frac{\sec ^{2} x}{5+3 \tan ^{2} x} d x=\int \frac{d t}{5+3 t^{2}}$

$=\frac{1}{3} \int \frac{1}{\frac{5}{3}+t^{2}} d t$

We know that $\int \frac{1}{\mathrm{a}^{2}+\mathrm{x}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$

$\Rightarrow \frac{1}{3} \int \frac{1}{\frac{5}{3}+t^{2}} d t=\frac{1}{3} \times \sqrt{\frac{5}{3}} \tan ^{-1}\left(\frac{t}{\sqrt{\frac{5}{3}}}\right)+c$

$=\frac{\sqrt{5}}{3 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{5}}\right)+c$

$\therefore \mathrm{I}=\int \frac{1}{3+2 \cos ^{2} \mathrm{x}} \mathrm{dx}=\frac{\sqrt{5}}{3 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} \tan \mathrm{x}}{\sqrt{5}}\right)+\mathrm{c}$

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