The integral _0^((x+3) x)/(1+3^2x),dx is equal to:

Question

The integral $\displaystyle\int_0^\pi\frac{(x+3)\sin x}{1+3\cos^2x}\,dx$ is equal to:

Accepted Answer

By King's rule ($x\to\pi-x$) and adding:

$$2I = \int_0^\pi\frac{(\pi+6)\sin x}{1+3\cos^2x}\,dx$$

With $t=\cos x$:

$$2I = (\pi+6)\int_{-1}^1\frac{dt}{1+3t^2} = \frac{\pi+6}{\sqrt3}\left[\tan^{-1}(\sqrt3\,t)\right]_{-1}^1 = \frac{\pi+6}{\sqrt3}\cdot\frac{2\pi}3$$

$$I = \frac{\pi(\pi+6)}{3\sqrt3}$$

Solutions · 1