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  • Evaluate the definite integrals in Exercises 25 to 33. 32.

Evaluate the definite integrals in Exercises 25 to 33.

    Q32.    \int_0^\pi\frac{x\tan x}{\sec x + \tan x} dx

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Let I = \int_0^\pi\frac{x\tan x}{\sec x + \tan x} dx               -(i)

Replacing x with (\pi -x),

\\ I = \int_\pi^0\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x) + \tan (\pi -x)} (-dx) \\ = -\int_\pi^0\frac{(\pi -x)(-)\tan x}{-\sec x - \tan x} dx

\\ \implies I = \int^\pi_0\frac{(\pi -x)\tan x}{\sec x + \tan x} dx                   - (ii)

Adding (i) and (ii)

 I + I = \int^\pi_0\left(\frac{x\tan x}{\sec x + \tan x} + \frac{(\pi -x)\tan x}{\sec x + \tan x} \right) dx

\implies 2I = \int^\pi_0\frac{\pi\tan x}{\sec x + \tan x} dx

\\ \implies 2I = \int^\pi_0\frac{\pi \frac{sin x}{cos x} }{\frac{1}{cos x} + \frac{sin x}{cos x}} dx \\ \implies 2I =\pi \int^\pi_0\frac{ sin x }{1+sin x} dx \\ \implies 2I =\pi \int^\pi_0\frac{ (1 +sin x ) -1}{1+sin x} dx \\ \implies 2I =\pi \int^\pi_0\left [1- \frac{1}{1+sin x} \right ]dx

\\ \implies 2I =\pi \int^\pi_0\left [1- \frac{1}{1+sin x} \right ]dx \\ \implies 2I =\pi \int^\pi_01 dx - \pi \int^\pi_0\frac{1}{1+sin x}.\frac{(1-sin x)}{(1 - sin x)}dx \\ \implies 2I =\pi\int^\pi_01 dx - \pi \int^\pi_0[\sec^2 x - \sec x \tan x]dx \\ \implies 2I =\pi[x]^\pi_0 - \pi[\sec x - \tan x]^\pi_0

\\ \implies 2I =\pi[\pi - 0] - \pi[tan \pi - sec \pi- tan \pi + sec 0] \\ \implies 2I =\pi[\pi -2] \\ \implies I =\frac{\pi}{2}[\pi -2]

Posted by

HARSH KANKARIA

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