Question

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

Answer 1

I = \[\int_0^\pi \frac{x \tan x}{secx + \tan x}dx = \int_0^\pi \frac{x sin x}{1 + \sin x}dx\] ...(i)

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

∴ I = \[\int_0^\pi \frac{\left( \pi - x \right) \sin\left( \pi - x \right)}{1 + \sin\left( \pi - x \right)}dx\]

⇒ I = \[\int_0^\pi \frac{\left( \pi - x \right) \sin x}{1 + sin x}dx\]

\[\int_0^\pi \frac{\pi \sin x - x \sin x}{1 + sinx}dx\]    ...(ii)

Adding (i) and (ii), we get:
2I = \[\int_0^\pi \frac{\pi \sin x}{1 + sin x}dx\]

⇒ 2I = \[\pi \int_0^\pi \frac{\sin x \left( 1 - \sin x \right)}{1 - \sin^2 x}dx\]

⇒ 2I = \[\pi \int_0^\pi \frac{\sin x - \sin^2 x}{\cos^2 x}dx\]

⇒ 2I =\[\pi \int_0^\pi \left( \text { tanx secx } - \tan^2 x \right) dx\]

⇒ 2I =\[\pi \int_0^\pi \left\{ \text { tanx secx} - \left( \sec^2 x - 1 \right) \right\} dx\]

⇒ 2I = \[\pi \int_0^\pi \left\{ \text { tanx secx } - \sec^2 x + 1 \right\} dx\]

⇒ 2I = \[\pi \left[ \text { secx - tanx } + x \right]_0^\pi\]

⇒ 2I = \[\pi \left[ \left( \sec \pi - \tan \pi + \pi \right) - \left( \sec 0 - \tan 0 + 0 \right) \right]\]

⇒ 2I = \[\pi \left[ \left( - 1 - 0 + \pi \right) - \left( 1 - 0 + 0 \right) \right]\]

⇒ 2I = \[\pi \left( \pi - 2 \right)\]

⇒ I = \[\frac{\pi}{2}  \left( \pi - 2 \right)\]

∴ \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x}dx\] = \[\frac{\pi}{2} \left( \pi - 2 \right)\].