How do you find the integral of int (sin(pix))^2*(cos(pix))^5 dx∫(sin(πx))2⋅(cos(πx))5dx?

1 Answer
Jim H
Oct 16, 2015

Add the form int sin^m u cos^n u du∫sinmucosnudu with at least one of m, nm,n odd to your mathematical cookbook.

Explanation:

int sin^m u cos^n u du∫sinmucosnudu with at least one of m, nm,n odd.
Integrate by substitution. Do this by pulling off one from the odd power, then convert the remaining even power to the other function. Integrate the resulting polynomial in sinusinu or cosucosu term by term.

I = int sin^2pixcos^5 pix dx = int sin^2pixcos^4 pix (cos pix )dx I=∫sin2πxcos5πxdx=∫sin2πxcos4πx(cosπx)dx

= int sin^2pix (underbrace(cos^2 pix)_"Replace")^2 (cos pix )dx

= int sin^2pix underbrace((1-sin^2 pix)^2)_"Expand" (cos pix )dx

= int sin^2pix (1-2sin^2 pix +sin^4 pix)(cos pix )dx

= int (sin^2pix -2sin^4 pix +sin^6 pix)(cos pix )dx

Let u = sinx so du = cosx dx and the integral becomes

= 1/pi int (u^2-2u^4+u^6)du

Finishing is left to the student.

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