Question

∫ sin³ x cos³ x dx

Answer 1

$\int \sin^3 x \cos ^3 x dx$

Let $t =\cos x$ , $dt = - \sin x dx$ . Then $\int \sin^3 x \cos ^3 x dx = \int -\sin^2x \, t^3 dt$

Also, $\sin^2x = 1 -\cos^2x = 1-t^2$ . So integral can be re-written as

$\displaystyle \int -(1-t^2) t^3 dt = - \int(t^3-t^5)dt = \frac{-t^4}{4} + \frac{t^6}{6} + C = \frac{\cos^6x}{6} - \frac{\cos^4x}{4}+C$

Answer: $\displaystyle \frac{\cos^6x}{6} - \frac{\cos^4x}{4}+C$