Question

Evaluate the following integral:

\[\int_0^\pi \left( \frac{x}{1 + \sin^2 x} + \cos^7 x \right)dx\]

Answer 1

\[\text{Let I }=\int_0^\pi \left( \frac{x}{1 + \sin^2 x} + \cos^7 x \right)dx ..................(1)\]

Then,

\[I = \int_0^\pi \left( \frac{\pi - x}{1 + \sin^2 \left( \pi - x \right)} + \cos^7 \left( \pi - x \right) \right)dx\]
\[ = \int_0^\pi \left( \frac{\pi - x}{1 + \sin^2 x} - \cos^7 x \right)dx ..................(2)\]

Adding (1) and (2), we get

\[2I = \int_0^\pi \left( \frac{x}{1 + \sin^2 x} + \cos^7 x + \frac{\pi - x}{1 + \sin^2 x} - \cos^7 x \right)dx\]
\[ \Rightarrow 2I = \pi \int_0^\pi \frac{1}{1 + \sin^2 x}dx\]

Dividing the numerator and denominator by cos²x, we get

\[2I = \pi \int_0^\pi \frac{\sec^2 x}{\sec^2 x + \tan^2 x}dx\]
\[ \Rightarrow 2I = \pi \int_0^\pi \frac{\sec^2 x}{1 + 2 \tan^2 x}dx\]
\[ \Rightarrow 2I = 2\pi \int_0^\frac{\pi}{2} \frac{\sec^2 x}{1 + 2 \tan^2 x}dx .....................\left[ \int_0^{2a} f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( 2a - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( 2a - x \right) = - f\left( x \right)\end{cases} \right]\]

Put tan x = z

Then

\[\sec^2 xdx = dz\]

When

\[x \to 0, z \to 0\]

When

\[x \to \frac{\pi}{2}, z \to \infty\]

\[\therefore 2I = 2\pi \int_0^\infty \frac{dz}{1 + \left( \sqrt{2}z \right)^2}\]
\[ \Rightarrow 2I = \left.2\pi \times \frac{\tan^{- 1} \sqrt{2}z}{\sqrt{2}}\right|_0^\infty \]
\[ \Rightarrow I = \frac{\pi}{\sqrt{2}}\left( \tan^{- 1} \infty - \tan^{- 1} 0 \right)\]
\[ \Rightarrow I = \frac{\pi}{\sqrt{2}} \times \left( \frac{\pi}{2} - 0 \right)\]
\[ \Rightarrow I = \frac{\pi^2}{2\sqrt{2}}\]