Question

What is the integral of `cos^5(x)`?

Answer 1

`=sinx+(sin^5x)/5-2/3*sin^3x+c`, where `c` is a constant

Explanation :

`=int(cos^5x) dx`

From trigonometric identity, which is

`cos^2x+sin^2x=1`, `=>cos^2x=1-sin2x`

`=int(cos^4x)*cos(x) dx`

`=int(cos^2x)^2*cos(x) dx`

`=int(1-sin^2x)^2*cos(x) dx` .. `(i)`

let's assume `sinx = t`, `=> (cosx) dx= dt`

substituting this in the `(i)`, we get

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

Now using expansion of `(1-y)^2=1+y^2-2y`, yields,

`=int(1+t^4-2t^2)dt`

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

`=t+t^5/5-2*t^3/3+c`, where `c` is a constant

`=t+t^5/5-2/3*t^3+c`, where `c` is a constant

now substituting `t` back gives,

`=sinx+(sinx)^5/5-2/3*(sinx)^3+c`, where `c` is a constant

`=sinx+(sin^5x)/5-2/3*sin^3x+c`, where `c` is a constant