Any thoughts on this integral? $\int \cos^2(x)\cdot\sin^4(x)dx$
I tried the usual trigonometric identities but they don't seem helpful
 A: $\int\cos^2(x)\sin^4(x)~dx=\frac14\int\sin^2(2x)\sin^2(x)~dx\\=\frac18\int\sin^2(2x)[1-\cos(2x)]~dx\\=\frac18\left[\int\sin^2(2x)~dx-\int\sin^2(2x)\cos(2x)~dx\right]$
Solve the first integral by writing $\sin^2(2x)=\frac12[1-\cos(4x)]$ and the second one by putting $\sin(2x)=t$.
A: Hint:
Use Intuition behind euler's formula
$$(2\cos x)^2(2i\sin x)^4=\left(e^{ix}+e^{-ix}\right)^2\left(e^{ix}-e^{-ix}\right)^4$$
If $2\cos(nx)=e^{inx}+e^{-inx}=u_n$
$$64\cos^2x\sin^4x=u_6+u_4(-4+2)+u_2(2+1-8+1+6)+u_0(-4)+12$$
A: Use $\cos(x)^2 = 1-\sin(x)^2$. Then you have a $\sin(x)^4$ and a $\sin(x)^6$ to integrate. These can either be computed by partial integration or by using trigonometric identities for powers of $\sin$.
A: Write $c$ for $\cos x$ and $s$ for $\sin x$. Then the integrand is 
\begin{align}
c^2 s^4 
&= c^2 s^2 (s^2) \\
&= \frac{1}{4} (2cs)^2 (1 - c^2) \\
&= \frac{1}{2} \frac{1}{4} (2cs)^2 (2 - 2c^2)  \\
&= \frac{1}{2} \left( \frac{1}{4} (2cs)^2 (1 - 2c^2)  \right) 
 + \frac{1}{2} \left( \frac{1}{4} (2cs)^2   \right) \\
\end{align}
Now $2cs = \sin 2x$, and $1 - 2c^2 = -\cos 2x$, so from here things should be relatively simple. 
A: You can lower the degree by noting that
$$
\cos^2x=\frac{1+\cos2x}{2},\qquad \sin^2x=\frac{1-\cos2x}{2}
$$
Thus you get
$$
\frac{(1-\cos^22x)(1-\cos2x)}{4}=\frac{1}{4}(1-\cos^22x-\cos2x+\cos^32x)=
\frac{1}{4}\left(1-\frac{1+\cos4x}{2}-\cos2x\sin^22x\right)
$$
A: Here's a cool thing. It's called a reduction formula.
Consider the integral
$$I(n)=\int \cos(x)^{n}\sin(x)^{2n}\mathrm dx$$
Then recall that $\sin(x)^2=1-\cos(x)^2$:
$$I(n)=\int \cos(x)^n\left(1-\cos(x)^2\right)^n\mathrm dx$$
Then assuming that $n$ is a non-negative integer, we recall the binomial formula:
$$(a-b)^n=\sum_{k=0}^{n}(-1)^k{n\choose k}a^{n-k}b^k$$
$$\left(1-\cos(x)^2\right)^n=\sum_{k=0}^{n}(-1)^k{n\choose k}\cos(x)^{2k}$$
So 
$$I(n)=\sum_{k=0}^{n}(-1)^k{n\choose k}\int\cos(x)^{2k+n}\mathrm dx$$
Then consider the integral 
$$C(m)=\int\cos(x)^m\mathrm dx$$
$$C(m)=\int\cos(x)^{m-1}\cos(x)\mathrm dx$$
IBP:
$$\mathrm dv=\cos(x)\mathrm dx\Rightarrow v=\sin(x)$$
$$u=\cos(x)^{m-1}\Rightarrow \mathrm du=-(m-1)\cos(x)^{m-2}\sin(x)\mathrm dx$$
So 
$$C(m)=\cos(x)^{m-1}\sin(x)+(m-1)\int\cos(x)^{m-2}\sin(x)^2\mathrm dx$$
$$C(m)=\cos(x)^{m-1}\sin(x)+(m-1)\int\cos(x)^{m-2}\mathrm dx-(m-1)\int\cos(x)^m\mathrm dx$$
$$C(m)=\cos(x)^{m-1}\sin(x)+(m-1)C(m-2)-(m-1)C(m)$$
$$mC(m)=\cos(x)^{m-1}\sin(x)+(m-1)C(m-2)$$
$$C(m)=\frac{\cos(x)^{m-1}\sin(x)}{m}+\frac{m-1}{m}C(m-2)$$
So 
$$I(n)=\sum_{k=0}^{n}(-1)^k{n\choose k}C(2k+n)$$
I know this isn't a very efficient method for too large to count on one hand, but it is still an alternate method. Because your integral is $I(2)$, we have that 
$$I(2)=\sum_{k=0}^{2}(-1)^k{2\choose k}C(2k+2)$$
$$I(2)=C(2)-2C(4)+C(6)$$
And from $C(m)=\frac{\cos(x)^{m-1}\sin(x)}{m}+\frac{m-1}{m}C(m-2)$ we see that 
$$I(2)=C(2)-2\left[\frac{\cos(x)^{3}\sin(x)}{4}+\frac{3}{4}C(2)\right]+\frac{\cos(x)^{5}\sin(x)}{6}+\frac{5}{6}C(4)$$
$$I(2)=-\frac12C(2)-\frac{\cos(x)^{3}\sin(x)}{2}+\frac{\cos(x)^{5}\sin(x)}{6}+\frac{5}{6}\left[\frac{\cos(x)^{3}\sin(x)}{4}+\frac{3}{4}C(2)\right]$$
$$I(2)=\frac18C(2)-\frac{7\cos(x)^{3}\sin(x)}{24}+\frac{\cos(x)^{5}\sin(x)}{6}$$
And since $C(0)=\int\mathrm dx=x$,
$$I(2)=\frac18\left[\frac{\cos(x)\sin(x)}{2}+\frac{x}{2}\right]-\frac{7\cos(x)^{3}\sin(x)}{24}+\frac{\cos(x)^{5}\sin(x)}{6}$$
$$I(2)=\frac{x}{16}+\frac{\cos(x)\sin(x)}{16}-\frac{7\cos(x)^{3}\sin(x)}{24}+\frac{\cos(x)^{5}\sin(x)}{6}$$
And there you go.
A: You could also use Euler’s Formula.
$$ \sin x = \frac{e^{ix} - e^{-ix }}{2i} $$
$$ \cos x = \frac{e^{ix} + e^{-ix}}{2} $$
$$\int {{(\frac{e^{ix} + e^{-ix }}{2})}^2} \bullet {(\frac{e^{ix} - e^{-ix }}{2i})}^4  dx $$
