# ∫〖Sin^(2 ) x cos2x dx〗 ( solving it, getting + c) could need some help

by Gillyjay
Tags: ∫&#sin2, cos2x, dx&#, solving
 P: 1 1. The problem statement, all variables and given/known data ∫〖Sin^(2 ) x * cos2x dx〗 2. Relevant equations The right answer should be: sin 2x / 4 - x/4 - sin4x / 16 3. The attempt at a solution If i would set Sin^(2 ) x= (1-cos2x)/2 I can replace it in the integral Then we get: ∫▒〖(1-cos2x)/2 cos2x dx〗 As far as I am concerned we cannot use the constant multiple rule because there is no constant to multiply I don’t know how to go on from here to solve it 
 PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,923 OK, so you have \begin{align*} \int \sin^2 x\cos 2x\,dx &= \int\left(\frac{1-\cos 2x}{2}\right)\cos 2x\,dx \\ &= \frac{1}{2}\int(\cos 2x - \cos^2 2x)\,dx \\ &= \frac{1}{2}\int\cos 2x\,dx - \frac{1}{2}\int \cos^2 2x\,dx \end{align*} Can you take it from there? Hint: use a trig identity for cos2 2x.

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