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

  • Thread starter Gillyjay
  • Start date
  • #1
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Homework Statement



∫〖Sin^(2 ) x * cos2x dx〗


Homework Equations



The right answer should be: sin 2x / 4 - x/4 - sin4x / 16

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 
 

Answers and Replies

  • #2
vela
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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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