I was just tutoring and a student's question was: Make a trig substitution to make evaluating $\int \cos^2(x)\sin(2x)dx$ simpler.
So yeah, the question is only asking for what trigonometric substitution do we use to simplify the integral, it's not asking to evaluate the integral.
FIRST ATTEMPT:
Using $\sin(2x)=2\cos(x)\sin(x)$ the integral becomes
$$\begin{aligned}&\int\cos^2(x)\sin(2x)dx\\&=\int\cos^2(x)2\cos(x)\sin(x)dx\\&=2\int\cos^3(x)\sin(x)\end{aligned}$$
I thought this was a good answer because if we set $u =\cos^2(x)$, then by the chain rule we get $$du = -2\cos(x)\sin(x)dx$$ and thus $$\frac{du}{-2\cos(x)\sin(x)}=dx$$
And so
$$\begin{aligned}&\int\cos^2(x)2\cos(x)\sin(x)dx\\&=2\int\cos^3(x)\sin(x)dx\\&=\int\cos^2(x)(2\cos(x)\sin(x))dx\\&=\int\cos^2(x)(2\cos(x)\sin(x))\left(\frac{du}{-2\cos(x)\sin(x)}\right)\\&=\int\cos^2(x)\\&=\int udu\end{aligned}$$
And is thus quite solvable. However, the computer said that this substitution was not correct.
SECOND ATTEMPT:
I used $\cos^2(x) = \frac{1+\cos(2x)}{2}$ to get:
$$\begin{aligned}\int\cos^2(x)\sin(2x)dx\\&=\int \left(\frac{1+\cos(2x)}2\right)\sin(2x)dx\\&=\int \left(\frac{\sin(2x)+\sin(2x)\cos(2x)}2\right)dx\\&=\int\left(\frac{sin(2x)}2+\frac{\sin(2x)\cos(2x)}2\right)dx\end{aligned}$$
which is quite a simplification, and we can even take this one further with realizing that since $\sin(2x)=2\sin(x)\cos(x)$ we have that $\sin(4x)=2\sin(2x)\cos(2x)$. Thus our integral becomes:
$$=\int\left(\frac{\sin(2x)}2+\frac{\sin(4x)}2\right)dx$$
which I think is quite simple. However, the homework program did not accept this substitution as the correct substitution to make either.
Does anyone have any ideas?
The simplest will be to put $$u=\cos(2x)$$ with
$$\cos^2(x)=\frac{1+\cos(2x)}2$$ and
$$du=-2\sin(2x)dx$$
the integral becomes
$$\int\frac{1+u}2\left(-\frac{du}2\right)=\frac{-1}4\left(u+\frac{u^2}2\right)$$