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

1 Answer
Jan 21, 2017

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!