Integral by substitution

Problem

Compute $\displaystyle\int_0^\pi x\sin(x^2)\,dx$ with the substitution $t=x^2$.

Solution

We set $t = x^2$, $dt = 2x\,dx$, so $x\,dx = \dfrac{dt}{2}$: $\int x\sin(x^2)\,dx = \frac{1}{2}\int\sin t\,dt = -\frac{1}{2}\cos t = -\frac{1}{2}\cos(x^2).$ Evaluating at the endpoints $x=0$ and $x=\pi$: $\int_0^\pi x\sin(x^2)\,dx = \left[-\frac{1}{2}\cos(x^2)\right]_0^\pi = \frac{1}{2}\bigl[1-\cos(\pi^2)\bigr] \approx \boxed{0{,}95}.$

Links

Topics: Integral
Concepts: Definite integral · Integral by substitution
Methods: Integration by substitution
Skills: Compute · Integrate
Exercise types: Integral computation