Evaluate the definite integral _ 0 ^ 2 ^ 2 x d x ^ 2 x+4 ^ 2 x

Question

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}$

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Answer

Given $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$
Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x ......(i)$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} \mathrm{x}}{\cos ^{2} \mathrm{x}+4\left(1-\cos ^{2} \mathrm{x}\right)} \mathrm{d} \mathrm{x}$
$=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4(1)-\left(4 \cos ^{2} x\right)} d x$
$ =\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{4-3 \cos ^{2} x} d x$
$ =\int_{0}^{\frac{\pi}{2}} \frac{\frac{1}{3} \cdot 3 \cos ^{2} x}{4-3 \cos ^{2} x} d x$
$ =-\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{-3 \cos ^{2} x+4-4}{4-3 \cos ^{2} x} d x$
$ =-\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4-3 \cos ^{2} x}{4-3 \cos ^{2} x} d x+\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4}{4-3 \cos ^{2} x} d x$
$ =-\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}}(1) d x+\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4}{4-3\left(\frac{1}{\\sec ^{2} x}\right)} d x$
$ =-\frac{1}{3} \cdot[\mathrm{x}]_{0}^{\frac{\pi}{2}}+\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4 \\sec ^{2} \mathrm{x}}{4 \\sec ^{2} \mathrm{x}-3} \mathrm{dx}$
$ =-\frac{1}{3} \cdot\left[\frac{\pi}{2}\right]+\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4 \\sec ^{2} x}{4\left(1+\tan ^{2} x\right)-3} d x$
$\Rightarrow \mathrm{I}=-\frac{\pi}{6}+\frac{2}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{2 \\sec ^{2} \mathrm{x}}{1+4 \tan ^{2} \mathrm{x}} \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=-\frac{\pi}{6}+\mathrm{I}_{1} .....(ii)$
First solve for $I_1:$
$I_{1}=\frac{2}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x$
Let $2 \tan x = t \Rightarrow 2 \sec^2x dx dt$
When $x = 0$ then $t = 0$ and when x $ =\frac{\pi}{2}$ then t $ =\infty$
$\Rightarrow \frac{2}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x=\frac{2}{3} \cdot \int_{0}^{\infty} \frac{1}{1+t^{2}} d t$
$\Rightarrow \mathrm{I}_{1}=\frac{2}{3}\left[\tan ^{-1} \mathrm{t}\right]_{0}^{\infty}$
$ =\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]$
$\Rightarrow \mathrm{I}_{1}=\frac{2}{3} \cdot \frac{\pi}{2}$
$\Rightarrow \mathrm{I}_{1}=\frac{\pi}{3}$
Put this value in equ.$(ii)$
$\Rightarrow \mathrm{I}=-\frac{\pi}{6}+\frac{\pi}{3}$
$\Rightarrow \mathrm{I}=\frac{\pi}{6}$

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