`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)`
...
`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.`
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