INTEGRALS — Maths STD 12 Science — Question

1. 1

​​​​​Solution:

We have, $\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\frac{\text{dx}}{1+\cos2\text{x}}=\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\frac{\text{dx}}{2\cos^2\text{x}}$ $[\because1+\cos2\text{A}=2\cos^2\text{A}]$

$=\frac{1}{2}\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\sec^2\text{x dx}$

$=\int\limits^{\frac{\pi}{4}}_0\sec^2\text{x dx}$  $\int\limits^\text{a}_{-\text{a}}\text{f(x)dx}=\begin{cases}\int\limits^\text{a}_0\text{f(x)},&\text{if f(x) is even}\\0,&\text{if f(x) is odd}\end{cases}$

$=\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_0=1$

$=\tan\frac{\pi}{4}-\tan0$

$=1$