Definite Integration Question - None

Question

If $f$ is continuous, prove/use: $\displaystyle\int_0^{\pi} x\,f(\sin x)\,dx = \frac{\pi}{2}\int_0^{\pi} f(\sin x)\,dx$. Hence evaluate $\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\cos^2 x}\,dx$

Accepted Answer

Step 1 (King): Let $I = \int_0^\pi \frac{x\sin x}{1+\cos^2 x}\,dx$. Sub $x\to\pi-x$:

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

Step 2: Add: $2I = \pi\int_0^\pi \frac{\sin x}{1+\cos^2 x}\,dx$

Step 3: Let $t=\cos x$, $dt=-\sin x\,dx$:

$$\int_0^\pi \frac{\sin x}{1+\cos^2 x}\,dx = \int_1^{-1}\frac{-dt}{1+t^2} = \int_{-1}^1\frac{dt}{1+t^2} = [\arctan t]_{-1}^1 = \frac{\pi}{4}-\left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$$

Step 4: $2I = \pi\cdot\dfrac{\pi}{2}\Rightarrow I = \boxed{\dfrac{\pi^2}{4}}$

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