How can I integrate (cosx)^3/sqrt(1+sinx)?

Question

Steps Please!

Answer 1

A good way is to get rid of that sqrt in the bottom. The usual way to do that is to utilize the identity: 1+ tan² = sec²
So, use tan² u = sin x
Then 2 tan u sec² u du = cos x dx
So now we want to integrate:
cos² x * 2 * tan u * sec² u * du/ sec u =
2cos²(x) * tan(u) * sec(u) du

But cos²(x) = 1-sin²(x) = (1-sin(x))(1+sin(x)) = (1-tan² u)(sec² u)
So we are integrating:
2(1-tan² u)* tan(u) sec³(u) du

Now the way to proceed is dictated by the fact that the derivative of the sec is sec * tan. That is to say, we would like only a single tan, but powers of sec's are OK.
Since tan² = sec² - 1 our integrand becomes:
2(2-sec²(u)) tan(u) sec³(u) du

The integral is thus:
(4/3)(sec(u))^3 -(2/5)(sec(u))^5 + C

Since sec²(u)-1 = sin(x)
sec(u) = sqrt(1+sinx) so that the final answer is:
(4/3)(1+sinx)^(3/2) - (2/5)(1+sinx)^(5/2) + C

Answer 2

You want to do a substitution u = ..., du = ... dx.

A good candidate would be the 1 + sin x under the sqrt sign. If u = 1 + sin x then du = cos x dx.

Aha! Replace cos^3 x with cos^2 x cos x = (1 - sin^2 x) cos x.

You get the integral of (1 - (u - 1)^2) u^(-1/2) du which is easy to do using integral u^k => (u^(k + 1)) / (k + 1) for k not= 1.

Dan