Question #301988

Evaluate the integral of sinΒ³ y cos 2y dy from 0 to πœ‹/2.

1
Expert's answer
2022-02-25T04:22:29-0500

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

∫0Ο€/2sin⁑3ycos⁑(2y)dy=βˆ’βˆ«10(1βˆ’u2)(2u2βˆ’1)du\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 ∫01\int_0^1 ) :

∫01(3u2βˆ’2u4βˆ’1)du=[u3βˆ’25u5βˆ’u]01=1βˆ’25βˆ’1=βˆ’25\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}


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