Question

How do you evaluate the integral `int cos^3xsqrtsinx`?

Answer 1

`2/3(sinx)^(3/2) - 2/7(sinx)^(7/2) + C`

Explanation:

Let `u = sinx`.

Then `du = cosxdx` and `dx = (du)/cosx`

`=>intcos^3xsqrt(u) * (du)/cosx`

`=>int cos^2x sqrt(u)du`

We can rewrite `cos^2x` as `1 - sin^2x`.

`=>int (1 - sin^2x)sqrt(u)du`

Since `u = sinx`, `u^2 = sin^2x`

`=>int(1 - u^2)sqrt(u)du`

`=> int(1 - u^2)u^(1/2)du`

`=>int u^(1/2) - u^(5/2)du`

This can be evaluate as `intx^ndx = x^(n + 1)/(n + 1) + C`

`=>2/3u^(3/2) - 2/7u^(7/2) + C`

`=>2/3(sinx)^(3/2) - 2/7(sinx)^(7/2) + C`

Hopefully this helps!