Question

What is `intcos^2(x)sin(x)dx`?

Answer 1

`intcos^2xsinxdx=-1/3cos^3x+c`

Explanation:

now`d/(dx)(cos^nx)`by the chain rule

`y=cos^nx`
`u=cosx=>(du)/(dx)=-sinx`

`d/(dx)(u^n)=n(u^(n-1))`

so

`dy/dx=(dy)/(du)xx(du)/(dx)`

`(dy)/(dx)=n(u^(n-1))xx(-sinx)=-ncos^(n-1)sinx`

so the integral

`intcos^2xsinxdx`

looks like a `y=cos^3x` function

`d/(dx)(cos^3x)=3cos^2x(-sinx)=-3cos^2xsinx`

reversing the process

`intcos^2xsinxdx=-1/3cos^3x+c`

Answer 2

`intcos^2(x)sin(x)dx=-cos^3x+c`

Explanation:

Let `u=cosx`, then `du--sinxdx`

Hence `intcos^2(x)sin(x)dx=-intu^2du`

`= -u^3/3+c`

`= -1/3cos^3x+c`