MCQ
$\int_0^x \frac{x \tan x}{\sec x+\tan x} d x=$
  • A
    $\frac{\pi}{2}-1$
  • B
    $\pi\left(\frac{\pi}{2}+1\right)$
  • C
    $\frac{\pi}{2}+1$
  • $\pi\left(\frac{\pi}{2}-1\right)$

Answer

Correct option: D.
$\pi\left(\frac{\pi}{2}-1\right)$
(D)
Let $I =\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ ...(i)
$\therefore \quad I=\int_0^\pi \frac{(\pi-x) \tan x}{\sec x+\tan x} d x$ ...(ii)
$\ldots .\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\pi \int_0^\pi \frac{\tan x}{\sec x+\tan x} d x$
$\therefore \quad I =\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\sin x} d x$
$=\frac{\pi}{2}\left[\int_0^\pi 1 d x-\int_0^\pi \frac{ d x}{1+\sin x}\right]$
On solving, we get
$I=\frac{\pi}{2}(\pi-2)=\pi\left(\frac{\pi}{2}-1\right)$

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