Question

How do you evaluate the indefinite integral `intsin^3(x)*cos^2(x)dx` ?

Answer 1

`=-(cos^3x)/3+(cos^5x)/5+c`, where `c` is a constant

Explanation :

`=intsin^3(x)*cos^2(x)dx`

`=intsin^2(x)*sin(x)*cos^2(x)dx`

from trigonometric identity `sin^2(x)+cos^2(x)=1`, we get `sin^2(x)=1-cos^2(x)`

`=intsin(x)*(1-cos^2(x))*cos^2(x)dx`

Integration by Substitution

let's `cos(x)=t`, `=>-sin(x)dx=dt`

`=int(1-t^2)*t^2(-dt)`

`=-int(1-t^2)*t^2dt`

`=-int(t^2-t^4)dt`

`=-intt^2dt+intt^4dt`

`=-t^3/3+t^5/5+c`, where `c` is a constant

substituting `t` back, yields

`=-(cos^3x)/3+(cos^5x)/5+c`, where `c` is a constant