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Q. The value of definite integral $\int\limits_0^\pi \frac{\pi \tan x}{\sec x+\tan x} d x$ is equal to

Integrals

Solution:

Let $I=\int\limits_0^\pi \frac{\pi \sin x}{1+\sin x} d x=2 \pi \int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{1+\sin x} d x=2 \pi \int\limits_0^{\frac{\pi}{2}} \frac{\cos x}{1+\cos x} d x$ (Using king property)
$\therefore \quad I=2 \pi \int\limits_0^{\frac{\pi}{2}} \frac{(1+\cos x)-1}{1+\cos x} d x=2 \pi \int\limits_0^{\frac{\pi}{2}}\left(1-\frac{1}{2 \cos ^2 \frac{x}{2}}\right) d x$
$=2 \pi \int\limits_0^{\frac{\pi}{2}}\left(1-\frac{1}{2} \sec ^2 \frac{x}{2}\right) d x=2 \pi\left(x-\tan \frac{x}{2}\right)_0^{\frac{\pi}{2}}=2 \pi\left(\frac{\pi}{2}-1\right)=\pi(\pi-2)$.