Integral Calculus
86413
The value of the integral \(\int_{0}^{\pi / 2} \frac{1}{1+\cos ^{2} x} d x\) is equal to
1 \(\frac{\pi}{4 \sqrt{2}}\)
2 \(\frac{\pi}{2}\)
3 \(\frac{\pi}{\sqrt{2}}\)
4 \(\frac{\pi}{2 \sqrt{2}}\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 2} \frac{1}{1+\cos ^{2} x} \cdot d x\)
\(=\int_{0}^{\pi / 2} \frac{\sec ^{2} x}{\sec ^{2} x+1} d x=\int_{0}^{\pi / 2} \frac{\sec ^{2} x}{\tan ^{2} x+2} d x\)
Put, \(\tan x=t\)
\(\sec ^{2} x \cdot d x=d t\)
When, \(x=0, t=0\) and \(x=\frac{\pi}{2}, t=\infty\)
\(\therefore \mathrm{I}=\int_{0}^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^{2}+(\sqrt{2})^{2}}=\frac{1}{\sqrt{2}}\left[\tan ^{-1}\left(\frac{\mathrm{t}}{\sqrt{2}}\right)\right]_{0}^{\infty}\)
\(=\frac{1}{\sqrt{2}}\left[\frac{\pi}{2}-0\right]=\frac{\pi}{2 \sqrt{2}}\)