Question

Evaluate the following integral:

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

Answer 1

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

Then,

\[I = \int_{- \pi}^\pi \frac{2x}{1 + \cos^2 x}dx + \int_{- \pi}^\pi \frac{2x\sin x}{1 + \cos^2 x}dx\]
\[ = I_1 + I_2\]

Consider

\[f\left( x \right) = \frac{2x}{1 + \cos^2 x}\]

Now,

\[f\left( - x \right) = \frac{2\left( - x \right)}{1 + \cos^2 \left( \pi - x \right)} = - \frac{2x}{1 + \left( - \cos x \right)^2} = - \frac{2x}{1 + \cos^2 x} = - f\left( x \right)\]

\[\therefore I_1 = \int_{- \pi}^\pi \frac{2x}{1 + \cos^2 x}dx = 0 \left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]

Again, consider

\[g\left( x \right) = \frac{2x\sin x}{1 + \cos^2 x}\]

\[g\left( - x \right) = \frac{2\left( - x \right)\sin\left( - x \right)}{1 + \cos^2 \left( - x \right)} = \frac{2x\sin x}{1 + \cos^2 x} = g\left( x \right) \left[ \sin\left( - x \right) = - \sin x and \cos\left( - x \right) = \cos x \right]\]

\[\therefore I_2 = \int_{- \pi}^\pi \frac{2x\sin x}{1 + \cos^2 x}dx\]
\[ = 2 \times 2 \int_0^\pi \frac{x\sin x}{1 + \cos^2 x}dx ................\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]
\[ = 4 \int_0^\pi \frac{x\sin x}{1 + \cos^2 x}dx ..................(1)\]

Then, 

\[I_2 = 4 \int_0^\pi \frac{\left( \pi - x \right)\sin\left( \pi - x \right)}{1 + \cos^2 \left( \pi - x \right)}dx = 4 \int_0^\pi \frac{\left( \pi - x \right)\sin x}{1 + \cos^2 x}dx .................(2) \left[ \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]

Adding (1) and (2), we get

\[2 I_2 = 4 \int_0^\pi \frac{\pi\sin x}{1 + \cos^2 x}dx\]
\[ \Rightarrow 2 I_2 = 4\pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x}dx\]

Put cos x = z

\[\Rightarrow - \sin x\ dx = dz\]

When

\[x \to 0, z \to 1\]

When 

\[x \to \pi, z \to - 1\]

\[\therefore 2 I_2 = - 4\pi \int_1^{- 1} \frac{dz}{1 + z^2}\]
\[ \Rightarrow 2 I_2 = - 4\pi \times \tan^{- 1} z_1^{- 1} \]
\[ \Rightarrow 2 I_2 = - 4\pi\left[ \tan^{- 1} \left( - 1 \right) - \tan^{- 1} 1 \right]\]
\[ \Rightarrow 2 I_2 = - 4\pi\left( - \frac{\pi}{4} - \frac{\pi}{4} \right) = 2 \pi^2 \]
\[ \Rightarrow I_2 = \pi^2\]

\[\therefore I = I_1 + I_2 = 0 + \pi^2 = \pi^2\]