trig integration

ijd5000

Junior Member
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Sep 3, 2013
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51
i have an integral with an even power of sine and cosine i'm looking for the general pattern how to solve these, i know it has to involve the half angel identities.

Heres a specific problem:

The function is: 3sin^2x cos^4 x dx from pi to 0.
 
The standard method is to replace using the half angle identities:

sin2(x)=1cos(2x)2\displaystyle \sin^2(x)=\dfrac{1-\cos(2x)}{2}

cos2(x)=1+cos(2x)2\displaystyle \cos^2(x)=\dfrac{1+\cos(2x)}{2}

so

sin2xcos4x=(1cos(2x)2)(1+cos(2x)2)2\displaystyle \sin^2x \cos^4x = \left(\dfrac{1-\cos(2x)}{2}\right)\left(\dfrac{1+\cos(2x)}{2} \right)^2 which is ugly, but doable. Repeated applications of the identity will be needed.

Alternatively:

sin2xcos4x=sin2x(1sin2x)2=sin2x2sin4x+sin6x\displaystyle \sin^2x \cos^4x = \sin^2x(1-\sin^2x)^2 = \sin^2x-2\sin^4x+\sin^6x

It might also be useful in some cases to use sinxcosx=12sin(2x)\displaystyle \sin x\cos x = \frac{1}{2}\sin(2x) for integrals such as

sin4xcos4xdx=116sin4(2x)dx\displaystyle \displaystyle \int \sin^4x\cos^4x dx = \frac{1}{16} \int \sin^4(2x) dx
 
Last edited:
Anotherway:

sin2(x) * cos4(x) = cos4(x) - cos6(x)

cos4(x) = 3/8 + 1/2 * cos(2x) + 1/8 * cos(4x)

cos6(x) = 5/16 + 1/32 * [cos(6x) + 6 * cos(4x) + 15 * cos(2x)]

and so on.....
 
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