Calculus 2 Trig Integrals with Cosine and Sine Derivatives Did I do the following problem correctly?  I searched online but couldn't find an answer so I'm asking here.  

$$I=\int \cos^2 \theta \sin (2\theta)\, \mathrm{d}\theta$$

My work: 

 A: You can do it easier. $$\int\cos^2\theta\sin2\theta d\theta=2\int\cos^3\theta\sin\theta d\theta=-\frac12\cos^4\theta+C$$
It's easy to see that this is equivalent to your formula, by plugging in $\cos^2\theta=1-\sin^2\theta$
A: It's correct! To check it yourself, differentiate it and try to manipulate it back into the original form. Note that you can leverage the double angle identity too, and claim
$$(\sin^2 \theta)' = 2 \sin \theta \cos \theta = \sin 2 \theta$$
Similarly,
$$(\sin^4 \theta)' = 4 \sin^3 \theta \cos \theta = 2 \sin^2 \theta \sin 2 \theta = 2(1 - \cos^2 \theta) \sin 2 \theta$$
Plug in these identities as you take the derivative and simplify as necessary.
A: It's much easier to do it as follows:\begin{align}\int\cos^2(\theta)\sin(2\theta)\,\mathrm d\theta&=2\int\cos^3(\theta)\sin(\theta)\,\mathrm d\theta\\&=\frac12\cos^4(\theta)\end{align}since $\int u^3\,\mathrm du=\frac14u^4$.
A: $$I=\int \cos^2 x \sin 2x dx= 2\int \cos^3 x \sin x dx-=2 \int t^3 dt= -2 \frac{t^4}{4}+C=-\frac{\cos^4 x}{2}+C.$$ We have use $\cos x=t$.
