The integral _0^ (8 x d x)/(4 ^2 x+ ^2 x) is equal to

Question

The integral $\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x}$ is equal to

Accepted Answer

$$\begin{aligned}
& I=\int_0^\pi \frac{8 x}{4 \cos ^2 x+\sin ^2 x} d x \ldots(1) \
& I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2(\pi-x)+\sin ^2(\pi-x)} d x \
& I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2 x+\sin ^2 x} d x \ldots(2)
\end{aligned}$$

Adding (1) and (2)

$$\begin{aligned}
& 2 I=8 \pi \int_0^\pi \frac{1}{4 \cos ^2 x+\sin ^2 x} d x \
& I=4 \pi \times 2 \int_0^\pi \frac{\sec ^2 x}{4 \tan ^2 x} d x
\end{aligned}$$

Put $\tan x=t$

$$\sec ^2 x d x=d t$$

$$I=8 \pi \int_0^{\infty} \frac{d t}{4+t^2}$$

$$\begin{aligned}
& I=8 \pi \frac{1}{2}\left(\tan ^{-1} \frac{t}{2}\right)_0^{\infty} \
& I=4 \pi\left(\frac{\pi}{2}\right) \
& I=2 \pi^2
\end{aligned}$$

Solutions · 1