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

1 Answer
Mar 21, 2018

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

Explanation:

#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#