Question

${\int }_0^{\pi }\frac{\tan x}{\sec x+\cos x}dx=$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. 0.0
• D. $\pi$

Answer 1

The correct option is B $\frac{\pi }{2}$

${\int }_0^{\pi }\frac{\tan x}{\sec x+\cos x}dx$
$={\int }_0^{\pi }\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\cos x}dx={\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx$
Put cos x = t
Then - sin x dx = dt
$x=0\Rightarrow t=\cos 0=1$
$x=\pi \Rightarrow t=\cos \pi =-1$

${\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx$
$={\int }_1^{-1}\frac{1}{1+t^2}(-dt)$
$={\int }_{-1}^1\frac{1}{1+t^2}dt$
$=\left[{\tan }^{-1}\right(t){]}_{-1}^1$
$={\tan }^{-1}\left(1\right)-{\tan }^{-1}(-1)$
$={\tan }^{-1}1+{\tan }^{-1}1$
$=2ta{n}^{-1}\left(1\right)$
$=2.\frac{\pi }{4}$
$=\frac{\pi }{2}$