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In solving the integral

$$\int\sin^5(2x)\cos^3(2x)~dx,$$

I misread it as

$$\int\sin^5(2x)\cos^3(\color{red}{3x})~dx.$$

After a while of failing to solve this, I realized my mistake. How would I go about solving the second integral?

On first glance I thought it could be integration by parts, but since there are no $\cos(2x)$ or $\sin(3x)$ terms that seems like the wrong approach. I thought of using the $\sin(x)\cos(y) = \dots$ and $\sin(x)\sin(y) = \dots$ identities, but that didn't pan out either.

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  • $\begingroup$ have you tried $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ (and corr. for $\sin$) and bashing it? Actually upon checking wolframalpha, I don't imagine that the coefficients would be so easy to find $\endgroup$ – Andres Mejia Sep 13 '18 at 3:35
  • $\begingroup$ You may write $\cos 3x=\cos (2x+x)$. $\endgroup$ – sirous Sep 13 '18 at 3:37
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Hint:

Use double and triple angle trigonometric identities and reduce them to the same angle:

$$\cos(3x)=4\cos^3(x)-3\cos(x)$$

$$\sin(2x)=2\sin(x)\cos(x)$$

Expanding them will lead you to multiple integrals involving different exponents of the $\sin(x)$ and $\cos(x)$ (their product) terms which becomes similar to your original problem which you have already solved.

Can you try it now?

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    $\begingroup$ I don't think I'm going to solve it explicitly, but I crunched it enough into the form $$\int \sin(x) p(\cos(x))~dx$$ where $p$ is some gross polynomial, which is good enough for me. Thanks! $\endgroup$ – Santana Afton Sep 13 '18 at 3:57
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One way is to use complex expressions.

Reexpress the integral as:

$$\int(\frac{e^{i2x} - e^{-i2x}}{2i})^5(\frac{e^{i3x} + e^{-i3x}}{2})^3 dx$$

then use binomial expansion and simplify as much as you can. Exponentials are easy to integrate, but some simplification will be necessary to recast it into a real form.

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