Integral/Powers of sine/Recursion/Example

A primitive function for the sine function ${}\sin x$ is ${}-\cos x$. In order to find a primitive function for ${}\sin ^{n}x$, we use integration by parts to get a recursive relation to a power with a smaller exponent. To make this more precise, we work over an interval, the primitive function shall start at ${}0$ and have the value ${}0$ there. For ${}n\geq 2$, with integration by parts, we get

${}{\begin{aligned}\int _{0}^{x}\sin ^{n}t\,dt&=\int _{0}^{x}\sin ^{n-2}t\cdot \sin ^{2}t\,dt\\&=\int _{0}^{x}\sin ^{n-2}t\cdot {\left(1-\cos ^{2}t\right)}\,dt\\&=\int _{0}^{x}\sin ^{n-2}t\,dt-\int _{0}^{x}{\left(\sin ^{n-2}t\cos t\right)}\cos t\,dt\\&=\int _{0}^{x}\sin ^{n-2}t\,dt-{\frac {\sin ^{n-1}t}{n-1}}\cos t|_{0}^{x}-{\frac {1}{n-1}}{\left(\int _{0}^{x}\sin ^{n}t\,dt\right)}.\end{aligned}}$

Multiplication with ${}n-1$ and rearranging yields

${}n\int _{0}^{x}\sin ^{n}t\,dt=(n-1)\int _{0}^{x}\sin ^{n-2}t\,dt-\sin ^{n-1}x\cos x\,.$

In particular, for ${}n=2$, we have

${}\int _{0}^{x}\sin ^{2}t\,dt={\frac {1}{2}}{\left(x-\sin x\cos x\right)}\,.$