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

by Gillyjay
Tags: ∫&#sin2, cos2x, dx&#, solving
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Gillyjay
#1
Jan25-11, 12:52 AM
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 donít know how to go on from here to solve it 
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vela
#2
Jan25-11, 01:14 AM
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OK, so you have

[tex]\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*}[/tex]

Can you take it from there?

Hint: use a trig identity for cos2 2x.


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