Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 5

int sin^2(pix)cos^5(pix) dx
To solve, apply the Pythagorean identity sin^2 theta + cos^2 theta =1 repeatedly until the integral is in the form int u^n du .
= int sin^2(pix)cos^3(pix)cos^2(pix) dx
=int sin^2(pix)cos^3(pix)(1-sin^2(pix)) dx
=int [sin^2(pix)cos^3(pix) - sin^4(pix)cos^3(pix)]dx
= int [ sin^2(pix)cos(pix)cos^2(pix) - sin^4(pix)cos(pix)cos^2(pix)]dx
= int[sin^2(pix)cos(pix)(1-sin^2(pix)) -sin^4(pix)cos(pix)(1-sin^2(pix))]dx
= int [sin^2(pix)cos(pix)-sin^4(pix)cos(pix) - sin^4(pix)cos(pix)+sin^6(pix)cos(pix)] dx
int [sin^2(pix)cos(pix)-2 sin^4(pix)cos(pix)+sin^6(pix)cos(pix)] dx
=intsin^2(pix)cos(pix)dx-int2sin^4(pix)cos(pix)dx+intsin^6(pix)cos(pix)dx
To take the integral of this, apply u-substitution method.
u = sin (pix)
du= pi cos (pix) dx
(du)/pi = cos(pix) dx
= int u^2 *(du)/pi - int 2u^4 * (du)/pi + intu^6 * (du)/pi
= 1/pi int u^2 du - 2/pi int u^4 du + int 1/pi u^6 du
= 1/pi*u^2/3-2/pi*u^5/5 + 1/pi*u^7/7 + C
= u^2/(3pi) - (2u^5)/(5pi) + u^7/(7pi) + C
And, substitute back u = sin (pix) .
= (sin^2 (pix))/(3pi) - (2sin^5(pix))/(5pi)+ (sin^7(pix))/(7pi) + C

Therefore, int sin^2(pix)cos^5(pix) dx= (sin^2 (pix))/(3pi) - (2sin^5(pix))/(5pi)+ (sin^7(pix))/(7pi) + C.