Computing Definite Integrals — Question 5

Question 5 - Solution

This integral suggests integration by parts because it is a product of $x$ and $\sin x$.

Let $$u=x, \qquad dv=\sin x\,dx.$$ Then $$du=dx, \qquad v=-\cos x.$$

Apply integration by parts: $$\int x\sin x\,dx = -x\cos x+\int \cos x\,dx = -x\cos x+\sin x.$$

Now evaluate from $0$ to $\pi$: $$\left[-x\cos x+\sin x\right]_{0}^{\pi}.$$

Compute each endpoint: $$-\pi\cos\pi+\sin\pi=\pi,$$ $$-0\cdot\cos 0+\sin 0=0.$$

Subtract: $$\pi-0=\pi.$$

$$\boxed{\pi}$$