Efficient Integration of sin^2(2t)cos^2(t)

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\displaystyle\int sin^22tcos^2t\ dt

This was part (b) to a question, the previous part of the question was to integrate \displaystyle\int sin^22tcost\ dt which I managed to do by expressing sin^22t as 4(sin^2t - sin^4t)

I tried a similar method for the integrand above, but didn't really go far with it. I'm not really sure what I'm trying to spot here, and I usually struggle with these integrals.
 
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You can try with \int sin^2(2t)dt - \int sin^2(2t)sin^2(t)dt, plus some trigonometric manipulations.

P.S.: Silly me, perhaps you only need some trigonometric manipulations.
 
hikaru1221 said:
You can try with \int sin^2(2t)dt - \int sin^2(2t)sin^2(t)dt, plus some trigonometric manipulations.

P.S.: Silly me, perhaps you only need some trigonometric manipulations.

thanks
 
phospho said:
\displaystyle\int sin^22tcos^2t\ dt

This was part (b) to a question, the previous part of the question was to integrate \displaystyle\int sin^22tcost\ dt which I managed to do by expressing sin^22t as 4(sin^2t - sin^4t)

I tried a similar method for the integrand above, but didn't really go far with it. I'm not really sure what I'm trying to spot here, and I usually struggle with these integrals.
Instead of changing the first factor, I'd use the identity ##\cos^2 t = \frac{1+\cos 2t}{2}## to get everything in terms of ##2t##. It should be pretty straightforward after that.

Your original approach would also work. You should check your textbook on how to handle integrals of the form
$$\int \cos^n x \sin^m x\,dx.$$ They're tedious, but there's a recipe to follow that depends on the evenness and oddness of ##m## and ##n##.
 
After following vela's advice, you may also find it useful to know cos(3x) = 4 cos3(x) - 3 cos(x)
 
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