What is the reduction formula for $\int\sin^nx \space dx$ from $0$ to $\pi$ or from $0$ to $2\pi$?

Question

I already know reduction formula for $\int_{0}^{\frac{\pi}{2}}\sin^nx \space dx$ and $\int_{0}^{\frac{\pi}{2}}\cos^nx \space dx$. They can be found here in this question.

However I am unable to find what would the formulas, when upper limit is changed to $\pi$ or $2\pi$. After tinkering in Wolfram, I have found some relations but unsure if they are correct and how to prove them. They are as follows:

$$ \begin{align*} \int_{0}^{\pi}\sin^nx \space dx &= 2\int_{0}^{\frac{\pi}{2}}\sin^nx \space dx \\ \int_{0}^{2\pi}\sin^nx \space dx &= \begin{cases} 0, & \mbox{ when $n$ is odd } \\ 4\int_{0}^{\frac{\pi}{2}}\sin^nx \space dx, & \mbox{ when $n$ is even}. \end{cases} \\ \int_{0}^{\pi}\cos^nx \space dx &= \begin{cases} 0, & \mbox{ when $n$ is odd } \\ 2\int_{0}^{\frac{\pi}{2}}\cos^nx \space dx, & \mbox{ when $n$ is even}. \end{cases} \\ \int_{0}^{2\pi}\cos^nx \space dx &= 2\int_{0}^{\pi}\cos^nx \space dx = \begin{cases} 0, & \mbox{ when $n$ is odd } \\ 4\int_{0}^{\frac{\pi}{2}}\cos^nx \space dx, & \mbox{ when $n$ is even}. \end{cases} \end{align*} $$

How to prove them?

Answer 1

You can use the relations $$ \sin(x+\pi/2)=\cos(x),\qquad \cos(x+\pi/2)=-\sin x\qquad (\star) $$ to express all the integrals of the form $\int_0^{k\pi/2}\cos^n x\ dx$ and $\int_0^{k\pi/2}\sin^n x\ dx$ as sums over integrals of the form $\int_0^{\pi/2}\cos^n x\ dx$ and $\int_0^{\pi/2}\sin^n x\ dx$ by breaking the integral as a sum of integrals taken over the intervals $[0,\pi/2],\ [\pi/2,\pi],\ \ldots,\ [(k-1)\pi/2,k\pi/2]$ and then using a change of variables of the form $x\mapsto x+\ell\pi/2$ to express each piece in terms of the original integrals $\int_0^{\pi/2}\sin^n x\ dx$ and $\int_0^{\pi/2}\cos^n x\ dx$.

The relations $(\star)$ follow from the addition formulas for $\sin$ and $\cos$, or if you are familiar with Euler's formula it can be derived by applying Euler's formula to the identity $ ie^{ix}=e^{i(x+\pi/2)} $ yielding $$ i\cos x-\sin x=\cos(x+\pi/2)+i\sin(x+\pi/2). $$ For example $$ \int_0^\pi \sin^n x\ dx=\int_0^{\pi/2}\sin^n x\ dx+\int_{\pi/2}^{\pi}\sin^n x\ dx=\int_0^{\pi/2}\sin^n x\ dx+\int_0^{\pi/2}\sin^n (x+\pi/2)\ dx $$ $$ =\int_0^{\pi/2}\sin^n x\ dx+\int_0^{\pi/2}\cos^n x\ dx. $$

Answer 2

This post shows a general formula and is here for reference purposes.

Let $\quad \int_0^{\pi/2}\sin^nx \space dx = \int_0^{\pi/2}\cos^nx \space dx = I_n$

Now, $$ \begin{align*} \int\limits_{0}^{k\pi/2}\sin^nx\space dx &= \int\limits_{0}^{\pi/2}\sin^nx\space dx + \int\limits_{\pi/2}^{2\pi/2}\sin^nx\space dx + \space \dots \space + \int\limits_{(k-1)\pi/2}^{k\pi/2}\sin^nx\space dx \\ &= \int\limits_{0}^{\pi/2} \sin^nx \space dx + \int\limits_{0}^{\pi/2} \sin^n \left(x+\frac{\pi}{2}\right) \space dx + \space \dots \space + \int\limits_{0}^{\pi/2} \sin^n \left(x+\frac{(k-1)\pi}{2}\right) \space dx \\ &= \int\limits_{0}^{\pi/2} \sin^nx \space dx + \int\limits_{0}^{\pi/2} \cos^nx \space dx + (-1)^n\int\limits_{0}^{\pi/2} \sin^nx \space dx + (-1)^n\int\limits_{0}^{\pi/2} \cos^nx \space dx + \space \dots \\ &= \big( 1+1+(-1)^n+(-1)^n+1+\dots \big)I_n \end{align*} $$

Similarly, we can show that $$ \int\limits_{0}^{k\pi/2}\cos^nx\space dx = \big( 1+(-1)^n+(-1)^n+1+1+\dots \big)I_n $$

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Therefore, the final relations are

When $n$ is even, $$ \int\limits_{0}^{k\pi/2}\sin^nx\space dx = \int\limits_{0}^{k\pi/2}\cos^nx\space dx = kI_n $$

When $n$ is odd and $m \in \{0,1,2,3,\dots \}$,

$$ \int\limits_{0}^{k\pi/2}\sin^nx\space dx = \begin{cases} \hfill 0 & \mbox{when $k=4m$} \\ \hfill I_n & \mbox{when $k=4m+1$} \\ 2I_n & \mbox{when $k=4m+2$} \\ \hfill I_n & \mbox{when $k=4m+3$} \\ \end{cases} \\ \int\limits_{0}^{k\pi/2}\cos^nx\space dx = \begin{cases} \hfill 0 & \mbox{when $k=4m$} \\ \hfill I_n & \mbox{when $k=4m+1$} \\ \hfill 0 & \mbox{when $k=4m+2$} \\ -I_n & \mbox{when $k=4m+3$} \\ \end{cases} $$ where $$ I_n=\begin{cases} \frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}...\frac{2}{3}, & \mbox{ when $n$ is odd } \\ \frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}...\frac{1}{2}\frac{\pi}{2}, & \mbox{ when $n$ is even}. \end{cases} $$