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I found in this link the reduction formula $$ \int_0^{\pi/2} \sin^m x \cos^n x \ dx = \begin{cases} \frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n)(m+n-2) \cdots 2} (\frac{\pi}{2}) & m, n \text{ even}\\[5pt] \frac{[(m-1)(m-3) \cdots (2\text{ or }1)][(n-1)(n-3) \cdots (2\text{ or }1)]}{(m+n)(m+n-2) \cdots (2\text{ or }1)} & \text{otherwise} \end{cases} $$

How can I prove it?

Michael Hardy
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Astrid A. Olave H.
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  • Do you know the Gamma Function? – Franklin Pezzuti Dyer Jun 27 '18 at 14:46
  • Integrate by parts on $\sin^mx\cos x$, which is immediate, to establish a recurrence relation. –  Jun 27 '18 at 14:48
  • @YvesDaoust I defined the integral as $I_{m,n}$ I find that $I_{m,n} = \frac{n-1}{m+1} I_{m+2,n-2}$ but I do not know how to use that – Astrid A. Olave H. Jun 27 '18 at 15:14
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    If you apply it several times, you will reduce to $n=1$, which is easy or $n=0$. –  Jun 27 '18 at 15:19
  • Since this pops up on Google, in terms of the Gamma function, this would simplify to $$\dfrac{\Gamma\left(\dfrac{m + 1}{2}\right) \Gamma\left(\dfrac{n + 1}{2}\right)}{2 \Gamma\left(\dfrac{m + n + 2}{2}\right)}.$$ One could prove this by noting that the expression I wrote above also satisfies the same recurrence. – Aryaman Maithani Sep 25 '23 at 22:15

2 Answers2

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Let $I_{m,n}=\int_0^{\pi/2} \sin^m x \cos^n x \ dx$, integrating by parts we find that $$ I_{m,n}=\frac{n-1}{m+1} I_{m+2,n-2} \qquad (1) \\ I_{m,n}=\frac{m-1}{n+1} I_{m-2,n+2} \qquad (2) $$

Using (1) when $n$ is odd, $$ I_{m,n}=\frac{(n-1)(n-3) \cdots 2}{(m+n-2) \cdots (m+1)}I_{m+n-1,1}=\frac{(n-1)(n-3) \cdots 2}{(m+n)(m+n-2) \cdots (m+1)} \quad (3) $$ Interchaging $m$ and $n$ in (3) we find $I_{m,n}$ when $m$ is odd.

Then $$ I_{m,n} = \frac{[(m-1)(m-3) \cdots (2\text{ or }1)][(n-1)(n-3) \cdots (2\text{ or }1)]}{(m+n)(m+n-2) \cdots (2\text{ or }1)} $$ when either $m$ or $n$ are odd.

If $n$ and $m$ are even

$$ I_{m,n}=\frac{(n-1)(n-3) \cdots 1}{(m+n-1) \cdots (m+1)}I_{m+n,0} $$

On the other hand, let $J_k=\int_0^{\pi/2} \sin^k x \ dx$. Again,integrating by parts we find: $$ J_k=\frac{k-1}{k}J_{k-2} $$

In particular when $k$ is even $$ J_k=\frac{(k-1)(k-3) \cdots 1}{k(k-2) \cdots 2}\left(\frac{\pi}{2}\right) $$

Then

$$ \begin{align} I_{m,n}&=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n-1)(m+n-3) \cdots 1} \left[\frac{(m+n-1)(m+n-3) \cdots 1}{(m+n)(m+n-2) \cdots 2} \left(\frac{\pi}{2}\right)\right]\\ &=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n)(m+n-2) \cdots 2}\left(\frac{\pi}{2}\right) \end{align} $$

Astrid A. Olave H.
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Hint-

break the expression into $\sin^{(m-1)} x \; \{ \cos ^n x \sin x\}$.
Let 2nd function which is in curly bracket and 1st function which is out side. Now integrate using integration by part.

G Cab
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Prabhat kumar
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