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Evaluate : π ∫ 0 X Tan X Sec X ⋅ C O S E C X D X .

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Question

Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .

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Solution

Given: \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] I = \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] =`int_0^x x sin^2` xdx          ...(1)

∵ \[\int_0^a f(x)dx = \int_0^a f(a - x)dx\]

From equation (1), we have:
I =

`int_0^x (π - x)sin ^2(π  - x) dx`

I =

`int_0^πsin^2 (π - x)dx - int_0^x x sin (π - x) dx`

\[\Rightarrow\] c

[From equation (1)]

\[\Rightarrow\] 2I 

`=int_0^x π sin^2 ` xdx

\[\Rightarrow\] I = \[\frac{\pi}{2} \int\limits_0^\pi \sin^2 xdx\] = \[\frac{\pi}{4} \int_0^\pi \left( 1 - \cos2x \right)dx = \frac{\pi}{4} \int_0^\pi dx - \frac{\pi}{4} \int_0^\pi \cos2xdx\]

\[\Rightarrow\] I

\[= \frac{\pi^2}{4} - \frac{\pi}{8} \left( \sin2x \right)_0^\pi = \frac{\pi^2}{4} - \frac{\pi}{8}\]`(sin 2π -0) = π^2/4.`

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2013-2014 (March) Delhi Set 2

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