Question

Integral of cos^5x*sin³xdx ?

Answer 1

`-1/6 cos^6x + 1/8 cos^8x + C`

Explanation:

We wish to solve the following: `int (cos^5x sin^3x) dx`. Since `cosx` is the derivative of `sinx`, this looks ripe for a u-substitution. However, we'd like there to be only one instance of `du`, so we need to figure out a way to transform the integral so that we only have a single power of `sin` or `cos`.

Luckily, this is easy. Recall the identity `sin^2x + cos^2x = 1`. From this, it follows that `sin^2x = 1 - cos^2x`. Using this fact, we can make the following changes:

`int (cos^5x sin^3x) dx`
`= int (cos^5x sin^2x sinx) dx`
`= int((cos^5x)(1 - cos^2x)(sinx))dx`

Now, let `u = cosx`. Then `du = -sinx`. This gives

`-int ( (u^5)(1 - u^2)) du`
`= -int (u^5 - u^7) du`
`= - (1/6 u^6 - 1/8 u^8) + C`
`= -1/6 cos^6x + 1/8 cos^8x + C`