Answer to Question #301988 in Calculus for Arvie

Question

Evaluate the integral of sin³ y cos 2y dy from 0 to 𝜋/2.

Answer 1

We will first use a substitution $\begin{cases} u&=\cos y \\ du&=-\sin y dy \end{cases}$ and use the fact that $\begin{cases} \sin^2 y = 1-\cos^2 y \\ \cos2y = 2\cos^2 y -1 \end{cases}$, which transforms the integral :

$\int_0^{\pi/2}\sin^3 y \cos(2y) dy=-\int_1^0 (1-u^2)(2u^2-1)du$

This integral can be easily calculated (we have used the minus sign before the integral to transform it into $\int_0^1$ ) :

$\int_0^1(3u^2-2u^4-1)du = [u^3-\frac{2}{5} u^5-u]^1_0=1-\frac{2}{5}-1=-\frac{2}{5}$