How do you find the integral of $sin^2(x)cos^6(x) dx$ ?

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Answer 1 · 2018-03-21T17:01:15

$(-1/8(cosx)^7+1/48(cosx)^5+5/192(cosx)^3)*sinx+5/256sin2x+5/1280x+C$

$int (sinx)^2*(cosx)^6*dx$

= $int [1-(cosx)^2]*(cosx)^6*dx$

= $int (cosx)^6*dx$ - $int (cosx)^8*dx$

After using $int (cosx)^n*dx=1/n(cosx)^(n-1)*sinx+(n-1)/nint(cosx)^(n-2)*dx$ reduction formula,

$int (sinx)^2*(cosx)^6*dx$

= $int (cosx)^6*dx$ - $int (cosx)^8*dx$

= $int (cosx)^6*dx$ - $[1/8(cosx)^7*sinx+7/8int (cosx)^6*dx]$

= $1/8int (cosx)^6*dx$ - $1/8(cosx)^7*sinx$

= $1/8*[1/6(cosx)^5*sinx+5/6int (cosx)^4*dx]$ - $1/8(cosx)^7*sinx$

= $5/48int (cosx)^4*dx$ + $1/48(cosx)^5*sinx$ - $1/8(cosx)^7*sinx$

= $5/48*[1/4(cosx)^3*sinx+3/4int (cosx)^2*dx]$ + $1/48(cosx)^5*sinx$ - $1/8(cosx)^7*sinx$

= $5/192(cosx)^3*sinx+5/64int (cosx)^2*dx$ + $1/48(cosx)^5*sinx$ - $1/8(cosx)^7*sinx$

= $5/192(cosx)^3*sinx+5/128int (1+cos2x)*dx$ + $1/48(cosx)^5*sinx$ - $1/8(cosx)^7*sinx$

= $5/192(cosx)^3*sinx+5/128*x+5/256sin2x$ + $1/48(cosx)^5*sinx$ - $1/8(cosx)^7*sinx+C$

How do you find the integral of sin^2(x)cos^6(x) dxsin^2(x)cos^6(x) dx?