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

[phospho]

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

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.

 

[hikaru1221]

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.

 

[phospho]

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

 

[vela]

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

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 $n$.

 

[haruspex]

After following vela's advice, you may also find it useful to know cos(3x) = 4 cos³(x) - 3 cos(x)