Integral of function from $0$ to $\pi$ using $\sin x=t$ substitution

Question

How to make substitution $sin(x) = t $ in integral of function $f(x)$ in general case, which is continuous on interval $ [0,\pi] $?

$\int_{0}^{\pi} f(x) dx$

Answer 1

You want to make the introduce the substitution $\phi(t) = x = \arcsin(t)$. The problem with this substitution is that the range of the function $\arcsin$ is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, which doesn't contain the integration interval $[0,\pi]$.

To get around this problem you should split the integral into two parts and you will have:

$$\int_0^{\pi}f(x)dx = \int_0^{\frac{\pi}{2}}f(x)dx + \int_{\frac{\pi}{2}}^{\pi}f(x)dx $$

For the first integral you can use substitution $x = \phi_1(t) = \sin^{-1} t$, where $\phi_1: [-1,1] \to \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ and for the second you should use $x = \phi_2(t) = \sin^{-1} t$, where where $\phi_2: [-1,1] \to \left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]$. Then you would have:

$$\int_0^{\pi}f(x)dx = \int_0^{\frac{\pi}{2}}f(x)dx + \int_{\frac{\pi}{2}}^{\pi}f(x)dx = \int_0^1 f(\phi_1(t))\phi_1'(t)dt + \int_1^0 f(\phi_2(t))\phi_2'(t)dt$$