Question

0^pi/4 sin^5 (x) dx

Answer 1

\[\large \int\limits_{0}^{\pi/4}\sin^5x \;dx\]

Hmm we have an `odd` power on the sine, that's very good.
That means we won't have to use the reduction formula.

Let's use a familiar identity, \(\large \sin^2\theta=1-\cos^2\theta\)
And convert 4 of the powers on sine to cosines.

Answer 2

\[\large \int\limits\limits_{0}^{\pi/4}\sin^5x \;dx \qquad = \qquad \int\limits\limits_{0}^{\pi/4}(\sin x)(\sin^2x)(\sin^2x) \;dx\]

Answer 3

\[\large \int\limits\limits\limits_{0}^{\pi/4}(\sin x)(1-\cos^2x)(1-\cos^2x) \;dx\]

From here we can apply a `U substitution`.
Let \(\large u=\cos x\)

Answer 4

See where we're going with this? :O

Answer 5

e9xcos(2x)dx