I was asked to solve
$$\int \cos^2x \sin 2x \space dx$$
I solved it using the identity $\cos^2x = \frac{1}{2}(1 + \cos(2x))$ and the substitution $u = 2x$.
$$\int \cos^2x \sin 2x \space dx = \frac{1}{2}\int(1+\cos2x)\sin2x \space dx$$
$$ \frac{1}{4} \int (\sin u + \sin u \cos u) \space du \tag{$u = 2x$}$$
$$= \frac{1}{4}(\cos u + \int t \space dt) \tag{$t = \sin u$}$$
$$=\frac{\cos 2x}{4} + \frac{\sin^2 2x}{8} + C$$
I can't find an error in my procedure. However, the correct answer is supposed to be $-\frac{1}{2}\cos^4 x + C$. I did the problem a few times to make sure I was not missing anything, and couldn't find a problem in my logic. Since there clearly is one, I was hoping someone could point it out to me and show me the correct approach to the problem.
Thanks in advance.
Best Answer
You should get:$$\int\sin(u)\,du =-\cos u+C=-\cos(2x)+C.$$ You lost the minus sign.
So your closed form, without your constant, is $$\frac{-\cos 2x}4+\frac{\sin^2 2x}8=\frac{-2\cos(2x)+1-\cos^2(2x)}8=\frac{2-(\cos(2x)+1)^2}{8}$$
Then substitute $\cos(2x)=2\cos^2x-1,$ and you get $$\frac{2-4\cos^4x}{8}=-\frac{\cos^2 x}{2}+\frac14,$$ which differs from the given answer by a constant.
So your approach works to give an equivalent correct answer, but you needed to fix the two integral you used.