I need to proof that $$\int_{0}^{\pi}xf(\sin(x))dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))dx$$ The problem says that the substitution $u=\pi-x$ is useful. Following their advice $u=\pi-x,du=-dx$ $$-\int_{0}^{\pi}(\pi-u)f(\sin(\pi-u))du$$ What do I do next, this the first time I encounter integrals writeen like this and I am not that familiar with proof thanks in advance !
Asked
Active
Viewed 90 times
0
-
You were told to substitute $x=\pi-u$ , not $x=\pi-4$. – Graham Kemp Mar 16 '18 at 04:19
-
1Also, the new integration limits are wrong. – Rócherz Mar 16 '18 at 04:22
2 Answers
5
\begin{align*} I&=\int_{0}^{\pi}(\pi-u)f(\sin(\pi-u))du\\ &=\int_{0}^{\pi}(\pi-u)f(\sin u)du\\ &=\pi\int_{0}^{\pi}f(\sin u)du-\int_{0}^{\pi}uf(\sin u)du\\ &=\pi\int_{0}^{\pi}f(\sin u)du-I. \end{align*}
user284331
- 56,315
0
$$\int_{0}^{\pi}xf(\sin(x))dx=$$ $$\int_{0}^{\pi}(\pi-x)f(\sin(\pi-x))dx=$$
$$\int_{0}^{\pi}(\pi-x)f(\sin x)dx=$$ $$\pi\int_{0}^{\pi}f(\sin x)du-\int_{0}^{\pi}xf(\sin x)dx \implies $$ $$\int_{0}^{\pi}xf(\sin(x))dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))dx$$
Mohammad Riazi-Kermani
- 69,891