I'm having a difficult time solving this integral. I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{1}{2}\int\sin x\cos x\sin4x\,dx$
After this step, I tried evaluating the integral by using the $\sin a\sin b$ property.
$\dfrac{1}{4}\sin^2x\sin4x + \dfrac{1}{4}\int\cos x(\cos5x-\cos3x)\,dx$
Hint: $\sin^2(x)\cos(4x)$ is an even function, hence it has a Fourier cosine series involving $\cos(nx)$ for $n\in\{0,1,2,3,4,5,6\}$. For any $n\in\mathbb{N}$, $$ \int \cos(nx)\,dx = C+\frac{\sin(nx)}{n}.$$ Actually: $$ \sin^2(x)\cos(4x) = -\frac{1}{4}\cos(2x)+\frac{1}{2}\cos(4x)-\frac{1}{4}\cos(6x).$$ Can you finish from here?
Use the fact that $ \cos 2x = \cos ^2 x - \sin^2 x = 1 - 2\sin^2 x $ , so $\sin^2 x = \frac{1 - \cos {2x}}{2}$
Replace it in your integral an it will get easy after spliting it into a few trivial. You will also have to use that $\cos\alpha\cos\beta=\frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)]$
If you prefer to do this with partial integration , you could use the fact that $ \sin x \cos x = \frac{1}{2} \sin 2x $, and using $ \sin\alpha\cos\beta=\frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)] $ you would also get two easy integrals.
$$\begin{align*} \sin^2 x \cos 4x &= \sin x (\sin x \cos 4x) \\ &= \frac{1}{2} \sin x (\sin 5x - \sin 3x) \\ &= \frac{1}{4}(\cos 4x - \cos 6x + \cos 4x - \cos 2x) \\ &= \frac{1}{2} \cos 4x - \frac{1}{4} \cos 6x - \frac{1}{4} \cos 2x. \end{align*}$$
Consequently, we immediately and rather trivially obtain $$\int \sin^2 x \cos 4x \, dx = \frac{1}{8} \sin 4x - \frac{1}{24} \sin 6x - \frac{1}{8} \sin 2x + C.$$
Alternatively, observe $$\begin{align*} \sin^2 x \cos 4x &= \sin^2 x (\cos 3x \cos x - \sin 3x \sin x) \\ &= \cos 3x \sin^2 x \cos x - \sin 3x \sin^3 x \\ &= \frac{1}{3} \left( \frac{d}{dx}\left[\sin^3 x\right] \cos 3x + \frac{d}{dx} \left[ \cos 3x \right] \sin^3 x \right) \\ &= \frac{1}{3} \frac{d}{dx}\left[\cos 3x \sin^3 x\right], \end{align*}$$ the last step due to the product rule applied to the functions $\sin^3 x$ and $\cos 3x$; thus $$\int \sin^2 x \cos 4x \, dx = \frac{1}{3} \cos 3x \sin^3 x + C.$$ It is quite straightforward to demonstrate that these antiderivatives are equivalent.
Here is the most straight way to solve this problem:
Notice that $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x $
So $\sin^2 x \cos 4x = \sin^2 (1-2\sin^2 2x) = \sin^2 [1-2(1-2\sin^2 x)^2] = -8\sin^6 x +8\sin^4 x -\sin^2 x$
By Reduction Formula: The integral can be easily solved.
Integrals of this sort can be done using some useful identities that are immediate corollaries of the angle-sum formulas $\cos A \cos B-\sin A \sin B=\cos (A+B)$ and $\sin A \cos B+\cos A \sin B=\sin (A+B),$ in conjunction with $\sin (-A)=-\sin A$ and $\cos (-A)=\cos A.$
We have $\cos (A+B)=\cos A \cos B -\sin A \sin B\;, $ and $\; \cos (A-B)=\cos A \cos B+\sin A \sin B.$ Adding these , we have $$\cos (A+B)+\cos (A-B)=2\cos A \cos B.$$ Subtracting them , we have $$\cos (A+B)-\cos (A-B)=-\sin A \sin B.$$ Similarly from the formulas for $\sin (A\pm B)$ we have $$\sin (A+B)+\sin (A-B)=2\sin A \cos B$$ and $$\sin (A+B)-\sin (A-B)=2\cos A \sin B.$$ From the angle-sum formulas, with $A=B$ and from $\cos^2A+\sin^2A=1,$ we have $$\sin^2 A=(1-\cos 2A)/2$$ and $$\cos^2 A=(1+\cos 2A)/2.$$ In particular $\sin^2 A \cos 4A=\frac {1}{2}(1-\cos 2A)\cos 4A=\frac {1}{2}(\cos 4A-\cos 2A\cos 4A)=\frac {1}{2}(\cos 4A-\frac {1}{2}(\cos 6A +\cos 2A)).$
For a lot more of this, see Trigonometry, by Hobson. (Dover Press re-print.)
Use following trigonometric identities to solve this question,
$\sin^2x=(\frac{1-\cos2x}{2})$
& $\cos A\cos B=\frac{\cos(A+B)+\cos(A-B)}{2}$.
Given integration $$I=\int \sin^2x\cos 4x\ dx$$ $$put\ \sin^2x=\frac{1-\cos 2x}{2}$$ $$I=\int \Big(\frac{1-\cos 2x}{2}\Big)\cos 4x\ dx$$ $$\Rightarrow I= \frac{1}{2}\int \Big(\cos 4x-\cos 2x \cdot \cos 4x\Big)dx$$ $$\Rightarrow I=\frac{1}{2}\int\Big(\cos 4x-\frac{\cos(2x+4x)+\cos(2x-4x)}{2}\Big)dx$$ $$\Rightarrow I=\frac{1}{2}\int\Big(\cos 4x-\frac{\cos 6x+cos(-2x)} {2}\Big)dx$$ $$\Rightarrow I=\frac{1}{2}\int\Big(\cos 4x-\frac{\cos 6x+\cos2x} {2}\Big)dx\ \ \ \ \ \ \ \ \ \ \because \cos(-\theta)=\cos\theta$$ $$\Rightarrow I=\frac{1}{2}\int \cos 4x\ dx-\frac{1}{4}\int \cos 6x\ dx-\frac{1}{4}\int \cos 2x\ dx$$ $$\Rightarrow I=\frac{1}{2}×\frac{\sin 4x}{4}-\frac{1}{4}×\frac{\sin 6x}{6}-\frac{1}{4}×\frac{\sin 2x}{2}+c$$ $$\Rightarrow I=\frac{\sin 4x}{8}-\frac{\sin 6x}{24}-\frac{\sin 2x}{8}+c$$
