Find `intsin^(6)x cosxdx`.

Explanation

[ Here, power of sin x is 5 which is an odd positive integer.
Therefore, put `z=cosx`.]
Let `z=cos x.` Then `dz= -sin x dx .` Now,
`intsin^(5)xdx int sin^(4)x sinx dx`
`=int(sin^(2)x)^(2)sinx dx`
`=int(1-cos^(2)x)^(2)sinx dx`
`=int(1-z^(2))^(2)(-dz) " " [ :' z=cosx]`
`= -int(1-2z^(2)+z^(4))dz`
`= -[z-2(z^(3))/(3)+(z^(5))/(5)]+c`
` = -z+(2)/(3)z^(3)-(z^(5))/(5)+c`
` = -cosx +(2)/(3)cos^(3)x-(cos^(5)x)/(5)+c`