`int sin^2(pi x) cos^5(pi x) dx` Evaluate the integral

`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.`