Question

How do you evaluate the integral of `x*(cos(x^2))^5`?

Answer 1

`sin^5(x^2)/10-sin^3(x^2)/3+sin(x^2)/2+C`

Explanation:

First, let

`u=x^2" "=>" "du=2xdx`

So, we have

`intxcos^5(x^2)dx`

Rearrange the `x` and `dx` and add a `2` to achieve `2xdx`. To balance the `2` inside the integral, multiply by `1//2` outside of the integral.

`=1/2intunderbrace(cos^5(x^2))_(cos^5(u))*underbrace(2xdx)_(du)`

Substituting, this equals

`=1/2intcos^5(u)du`

Before we integrate this, we want to modify this so that we can have a `v` and `dv` term both in the integral. We can do this by writing `cos^5(u)` as

`cos^5(u)=cos(u)cos^4(u)=cos(u)(cos^2(u))^2=cos(u)(1-sin^2(u))^2`

Giving us the integral

`=1/2intcos(u)(1-sin^2(u))^2du`

Now, we can substitute again:

`v=sin(u)" "=>" "dv=cos(u)du`

Rearrange the integral:

`=1/2intunderbrace((1-sin^2(u))^2) _ ((1-v^2)^2) * underbrace(cos(u)du) _ (dv)`

Which equals:

`=1/2int(1-v^2)^2dv`

Distribute `(1-v^2)^2`.

`=1/2int(1-2v^2+v^4)dv`

Splitting up the integral, we obtain:

`=1/2intv^4dv-2(1/2)intv^2dv+1/2int1dv`

Using the rule:

`intv^ndv=v^(n+1)/(n+1)+C`

Yields, after integrating term by term:

`=1/2(v^5/5)-v^3/3+1/2(v)+C`

`=v^5/10-v^3/3+v/2+C`

Using `v=sin(u)`:

`=sin^5(u)/10-sin^3(u)/3+sin(u)/2+C`

Now using `u=x^2`:

`=sin^5(x^2)/10-sin^3(x^2)/3+sin(x^2)/2+C`