What is the integral of sin^5(x)dx?

Answer 1

∫sin⁵ x dx

First, extract a factor of sin x:

∫sin⁴ x sin x dx

Now, use the Pythagorean theorem:

∫(1-cos² x)² sin x dx

Expand:

∫sin x - 2 cos² x sin x + cos⁴ x sin x dx

Make the substitution u=cos x, du=-sin x dx:

∫-1 + 2u² - u⁴ du

Integrate:

-u + 2u³/3 - u⁵/5 + C

Resubstitute:

- cos x + 2/3 cos³ x - 1/5 cos⁵ x + C.

And we are done.

Answer 2

sin^2(x) = 1 - cos^2(x)
sin^5(x) = (1 - cos^2(x))^2 sin(x)

Let u = cos(x)
then du = - sin(x)

int(sin^5(x)dx) = int((u^2 - 1)^2 du)
= int(u^4 - 2u^2 + 1)du
= u^5/5 - 2u^3/3 + u

Replacing, the value of the integral is

cos^5(x)/5 - 2u cos^3(x)/3 + cos(x) + C