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AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 78 of 160

Marks: +1, -0

\int\limits_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x =

$\frac{\pi (\pi-2)}{2}$
$\frac{\pi+2}{2}$
$\frac{\pi (\pi+2)}{2}$
$\frac{\pi-2}{2}$

Solution:

\int\limits_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x

Let

I = \int\limits_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x

...(i)

\because \int\limits_{a}^{b} f(x) d x = \int\limits_{a}^{b} f(a+b-x) d x

\Rightarrow I = \int\limits_{0}^{\pi} \frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)} d x

I = \int\limits_{0}^{\pi} \frac{(\pi-x)(-\tan x)}{-\sec x-\tan x} d x

I = \int\limits_{0}^{\pi} \frac{(\pi-x) \tan x}{\sec x+\tan x} d x

...(ii)

Adding Eqs. (i) and (ii), we get

2 I = \int\limits_{0}^{\pi} \frac{x \tan x+(\pi-x) \tan x}{\sec x+\tan x} d x

2 I = \int\limits_{0}^{\pi} \frac{\pi \tan x}{\sec x+\tan x} d x

I = \frac{\pi}{2} \int\limits_{0}^{\pi} \frac{\tan x}{\sec x+\tan x} d x = \frac{\pi}{2} \int\limits_{0}^{\pi} \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} d x

= \frac{\pi}{2} \int\limits_{0}^{\pi} \left( \frac{\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} \right) d x

= \frac{\pi}{2} \int\limits_{0}^{\pi} \frac{\sin x(1-\sin x)}{1-\sin^{2} x} d x

= \frac{\pi}{2} \int\limits_{0}^{\pi} \frac{\sin x - \sin^{2} x}{\cos^{2} x} d x

I = \frac{\pi}{2} \int\limits_{0}^{\pi} \left( \frac{\sin x}{\cos^{2} x} - \frac{\sin^{2} x}{\cos^{2} x} \right) d x

\frac{\pi}{2} \int\limits_{0}^{\pi} \left( \sec x \tan x - \tan^{2} x \right) d x

\frac{\pi}{2} \int\limits_{0}^{\pi} \left\{ \sec x \tan x - (\sec^{2} x - 1) \right\} d x

I = \frac{\pi}{2} \int\limits_{0}^{\pi} \left( \sec x \tan x - \sec^{2} x + 1 \right) d x

= \frac{\pi}{2} \left[ \sec x - \tan x + x \right]_{0}^{\pi}

= \frac{\pi}{2} \left\{ (\sec \pi - \tan \pi + \pi) - \frac{\pi}{2} \right\}

\left\{ \sec 0^{\circ} - \tan 0^{\circ} + 0 \right\}

= \frac{\pi}{2} \left\{ (-1-0+\pi) - (1-0+0) \right\}

= \frac{\pi}{2} (\pi-1-1) = \frac{\pi}{2} (\pi-2)

I = \frac{\pi (\pi-2)}{2}

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