My Math Forum Evaluation of (sin x)^m (cos x)^n dx
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 August 17th, 2012, 10:54 AM #1 Senior Member   Joined: Aug 2012 From: New Delhi, India Posts: 157 Thanks: 0 Evaluation of (sin x)^m (cos x)^n dx Given the values of m & n how would you evaluate $I=\int_{0}^{\pi/2}\sin^{\small{m}}{x} \cos^{\small{n}}{x}\,dx$ without means of Gamma function?
 August 17th, 2012, 11:49 AM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Evaluation of (sin x)^m (cos x)^n dx Just a Hint I would first rewrite is like this $\int^{\frac{\pi}{2}}_{0} \sin^{m-1}(x) \sin(x) \cos^{n}(x) \,\, dx$ using integrating by parts $u=\sin^{m-1}(x) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, du=(m-1)\sin^{m-2}(x) \cos(x) \, dx$ $dv= \sin(x) \cos^{n}(x) \, dx \,\,\,\,\,\,\,\,\, v=-\frac {\cos^{n+1}(x)}{n+1}$ $I=\frac{m-1}{n+1} \int^{\frac{\pi}{2}}_{0}\cos^{n+2}(x) \sin^{m-1}(x)$ I would continue until I get sin(x)
August 17th, 2012, 01:41 PM   #3
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Re: Evaluation of (sin x)^m (cos x)^n dx

Hello, Rejjy!

Quote:
 $\text{Given the values of }m\text{ and }n\text{ how would you evaluate:} \;\;\;I \:=\:\int_{0}^{\;\;\;\;\pi/2}\sin^{^m}\!x\,\cos^{^n}\!x\,dx \text{without means of Gamma function?}$

$\text{If either }m\text{ or }n\text{ is odd, factor out one factor of that function to use as }du
\;\;\;\text{Convert the rest into the "other function".$

$\text{Example: }\:\int \sin^{^2}x\,\cos^{^5}x\,dx$

$\text{Since }n= 5\text{ is odd, factor out a }\cos\,\!x$

$\text{W\!e have: }\:\int\sin^{^2}x\,\cos^{^4}x\,(\cos\,\!x\,dx)$

$\text{Convert the rest to }\sin\,\!x:$

[color=beige]. . [/color]$\int \sin^{^2}x(\cos^{^2}x)^{^2}(\cos\,\!x\,dx)$

[color=beige]. . [/color]$=\;\int\sin^{^2}x(1\,-\,\sin^{^2}x)^{^2}(\cos\,\!x\,dx)$

[color=beige]. . [/color]$=\;\int\sin^{^2}x(1\,-\,2\sin^{^2}x\,+\,\sin^{^4}x)(\cos\,\!x\,dx)$

[color=beige]. . [/color]$=\;\int(\sin^{^2}x\,-\,2\sin^{^4}x\,+\,\sin^{^6}x)(\cos\,\!x\,dx)$

$\text{Let }u \,=\,\sin x\;\;\;\Rightarrow\;\;\;du\,=\,\cos\,\!x\,dx$

$\text{And we have: }\;\int(u^{^2}\,-\,2u^{^4}\,+\,u^{^6})\,du
\;\;\;\text{which is easily integrated.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\text{If both }m\text{ and }n\text{ are even,}
\;\;\;\text{we must use the Double-angle Identities:}
\;\;\;\;\;\sin^{^2}x \:=\:\frac{1\,-\,\cos2x}{2}\;\;\;\;\;\cos^{^2}x \:=\:\frac{1\,+\,\cos2x}{2}$

$\text{Example: }\;\int\sin^{^2}x\,\cos^{^2}x\,dx$

$\text{W\!e have: }\;\int\left(\frac{1\,-\,\cos2x}{2}\right)\left(\frac{1\,+\,\cos2x}{2}\ri ght)\,dx \;=\;\frac{1}{4}\int(1\,-\,\cos^{^2}2x)\,dx$

[color=beige]. . . . . . [/color]$=\;\frac{1}{4}\int\sin^{^2}2x\,dx \;=\;\frac{1}{4}\int\left(\frac{1\,-\,\cos4x}{2}\right)\,dx$

[color=beige]. . . . . . [/color]$=\;\frac{1}{8}\int(1\,-\,\cos4x)\,dx \;=\;\frac{1}{8}\,\left(x\,-\,\frac{1}{4}\,\!\sin4x\right)\,+\,C$

 August 18th, 2012, 01:19 AM #4 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Evaluation of (sin x)^m (cos x)^n dx What about something along the lines of m=7/2 and n=3/2.
 August 18th, 2012, 07:46 AM #5 Senior Member   Joined: Aug 2012 From: New Delhi, India Posts: 157 Thanks: 0 Re: Evaluation of (sin x)^m (cos x)^n dx And what if m and n are very large? I have a book that states the following result. $I=\frac{[(m-1)(m-3)(m-5)...2or1][(n-1)(n-3)....2or1]}{(m+n)(m+n-2)...2or1}$ Also, it says to multiple the above by $\pi/2$ if both m and n are even. @The_Fool Agreed, that is an interesting question. However, I am not aware if the above result holds for fractional m & n. It seems like it won't. Regards, Rejjy 18-Aug-2012 21:16 IST
August 18th, 2012, 09:10 AM   #6
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Re: Evaluation of (sin x)^m (cos x)^n dx

Quote:
 Originally Posted by The_Fool What about something along the lines of m=7/2 and n=3/2.
If my calculations are correct (I'm pretty sure they are correct) :

$\int_{0}^{\frac{\pi}{2}} \sin^{\frac{7}{2}}(x) \cos^{\frac{3}{2}}(x) dx= \frac{\sqrt{2}}{12} \text{K}(2) + \frac{i }{3 \sqrt{\pi}} \cdot \Gamma$$\frac{5}{4}$$^2$

where $K$ is the complete elliptic integral of the first kind and $\Gamma$ is the gamma function.

Balarka

.

 August 18th, 2012, 09:22 AM #7 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Evaluation of (sin x)^m (cos x)^n dx Now for the original problem, I would prefer zaidalyafey's method : if $I(m,n)= \int_{0}^{\frac{\pi}{2}} \sin(x)^m \cos(x)^n dx$ then, $I(m,n)= \frac{m-1}{n+1} I(m-1,n+2) = \frac{m-1}{n+1} \cdot \frac{m-2}{n+3} I(m-2,n+4) = \frac{m-1}{n+1} \cdot \frac{m-2}{n+3} \cdot \frac{m-3}{n+5} \cdot \, \cdot \, \cdot$ $= \prod_{k=1}^{m} \frac{m-k}{n-(2k-1)} I(0,2k)$ where $I(0,2k)= \int_{0}^{\frac{\pi}{2}} \cos^{2k} (x) dx$ which is elementary and easy to calculate. Balarka .
August 18th, 2012, 02:24 PM   #8
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Re: Evaluation of (sin x)^m (cos x)^n dx

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by The_Fool What about something along the lines of m=7/2 and n=3/2.
If my calculations are correct (I'm pretty sure they are correct) :

$\int_{0}^{\frac{\pi}{2}} \sin^{\frac{7}{2}}(x) \cos^{\frac{3}{2}}(x) dx= \frac{\sqrt{2}}{12} \text{K}(2) + \frac{i }{3 \sqrt{\pi}} \cdot \Gamma$$\frac{5}{4}$$^2$

where $K$ is the complete elliptic integral of the first kind and $\Gamma$ is the gamma function.

Balarka

.
But, you used the gamma function. The original post asked for a way to evaluate it without the gamma function.

August 18th, 2012, 10:22 PM   #9
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Re: Evaluation of (sin x)^m (cos x)^n dx

Quote:
 Originally Posted by The_Fool But, you used the gamma function. The original post asked for a way to evaluate it without the gamma function.
It's impossible to do it without any type of special function for $m,n \cancel{\in} \mathbb{Z^{+}}$

 August 19th, 2012, 01:16 AM #10 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Evaluation of (sin x)^m (cos x)^n dx Well, I don't think that's right anyway since that integral is real: $\int_{0}^{\pi/2}$$\sin x$$^{7/2}$$\cos x$$^{3/2}dx=\frac{1}{2}\beta(\frac{9}{4},\frac{5}{4})$ I used the beta function instead of the gamma function. While beta can be defined by gamma, it can stand on it's own without it in several other forms.

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