Ιntegral calculation : (sin(x))^4 * (cos(x))^6

  • #1
Summary: Ιntegral calculation : (sin(x))^4 * (cos(x))^6

Hi all,
I tried to solve it, but I got stuck. An advice from my professor is to set: x=arctan(t)

Τhanks.
New Doc 2019-09-26 18.01.07.jpg
 

Answers and Replies

  • #2
PeroK
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Alternatively, you can get a formulas for the integral of ##\sin^n x## and ##\cos^n x## from repeated integration by parts.

They are on my list of standard integrals.
 
  • #3
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Since both trigonometric functions lead to one another and back by differentiation and integration, integration by parts is usually how they are solved. Doing it twice should reduce the power of one while keeping the other one stable. This leads to a recursion, or can just be repeated until a single function is left.

I have found the formula on Wikipedia.
 
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  • #4
Thanks for your answers! I tried another way using the following trigonometric transformation formulas:

New Doc 2019-09-26 19.48.48_3.jpg


Now it's in simple format:
New Doc 2019-09-26 19.48.48_1.jpg


New Doc 2019-09-26 19.48.48_2.jpg


Do you know a better - faster way to solve it ?

Thanks.
 
  • #5
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Yes, integration by parts. Still.
$$
\int \sin^n x \cos^m x \,dx = -\dfrac{\sin^{n-1}x \cos^{m+1}x}{n+m} +\dfrac{n-1}{n+m}\int \sin^{n-2} x \cos^m x \,dx
$$
 
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  • #6
PeroK
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Do you know a better - faster way to solve it ?

Thanks.
I would have started with ##\sin^4 x = (1 - \cos^2 x)^2 = 1 - 2\cos^2x + cos^4x##.

Then used parts to calculate the integral of the various powers of ##\cos x##.

This is so useful and common that I keep the recursion formula in my list of standard integrals.
 
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  • #7
PeroK
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Yes, integration by parts. Still.
$$
\int \sin^n x \cos^m x \,dx = -\dfrac{\sin^{n-1}x \cos^{m+1}x}{n+m} +\dfrac{n-1}{n+m}\int \sin^{n-2} x \cos^m x \,dx
$$
A thing of beauty!
 
  • #8

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