Question

Evaluate the following integrals:

${\int }_0^{\mathrm{\pi }}\left(\frac{x}{1+{\sin }^{2}x}+{\cos }^{7}x\right)dx$

Answer 1

Let I = ${\int }_0^{\mathrm{\pi }}\left(\frac{x}{1+{\sin }^{2}x}+{\cos }^{7}x\right)dx$ .....(1)
Then,

$$\begin{gathered} I={\int }_0^{\mathrm{\pi }}\left(\frac{\mathrm{\pi }-x}{1+{\sin }^2\left(\mathrm{\pi }-x\right)}+{\cos }^7\left(\mathrm{\pi }-x\right)\right)dx \\ ={\int }_0^{\mathrm{\pi }}\left(\frac{\mathrm{\pi }-x}{1+{\sin }^{2}x}-{\cos }^{7}x\right)dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_0^{\mathrm{\pi }}\left(\frac{x}{1+{\sin }^{2}x}+{\cos }^{7}x+\frac{\mathrm{\pi }-x}{1+{\sin }^{2}x}-{\cos }^{7}x\right)dx \\ \Rightarrow 2I=\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{1}{1+{\sin }^{2}x}dx \end{gathered}$$

Dividing the numerator and denominator by cos²x, we get

$$\begin{gathered} 2I=\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{{\sec }^{2}x}{{\sec }^{2}x+{\tan }^{2}x}dx \\ \Rightarrow 2I=\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{{\sec }^{2}x}{1+2{\tan }^{2}x}dx \\ \Rightarrow 2I=2\mathrm{\pi }{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^{2}x}{1+2{\tan }^{2}x}dx\left[{\int }_0^{2a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^{a}f\left(x\right)dx,& \mathrm{if}f\left(2a-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(2a-x\right)=-f\left(x\right)\end{array}\right.\right] \end{gathered}$$
Put tanx = z

Then ${\sec }^{2}xdx=dz$

When $x\to 0,z\to 0$

When $x\to \frac{\mathrm{\pi }}{2},z\to \infty$

$$\begin{gathered} \therefore 2I=2\mathrm{\pi }{\int }_0^{\infty }\frac{dz}{1+{\left(\sqrt{2}z\right)}^2} \\ \Rightarrow 2I=2\mathrm{\pi }\times {\overline{)\frac{{\tan }^{-1}\sqrt{2}z}{\sqrt{2}}}}_0^{\infty } \\ \Rightarrow I=\frac{\mathrm{\pi }}{\sqrt{2}}\left({\tan }^{-1}\infty -{\tan }^{-1}0\right) \\ \Rightarrow I=\frac{\mathrm{\pi }}{\sqrt{2}}\times \left(\frac{\mathrm{\pi }}{2}-0\right) \\ \Rightarrow I=\frac{{\mathrm{\pi }}^2}{2\sqrt{2}} \end{gathered}$$