Evaluate $\int \cos^2x\sin^4x\mathrm{d}x$ 
Evaluate integral $$\int \cos^2x\sin^4x\mathrm{d}x.$$

Attempt. Setting $\tan x=t$, gives:
$$\int \cos^2x\sin^4x\mathrm{d}x =\int \frac{1}{1+t^2} \,\left(\frac{t^2}{1+t^2}\right)^2 \frac{\mathrm{d}t}{1+t^2}=\int \frac{t^4}{(1+t^2)^4} \mathrm{d}t,$$
which does not seem to be elementary.
Thank in advance for the help.
 A: Use that $$\cos^2(x)=1-\sin^2(x)$$ and your integral will be
$$\int\sin^4(x)-\sin^6(x)dx$$
For your work:
$$\int \sin(x)^4dx=-1/4\, \left(  \left( \sin \left( x \right)  \right) ^{3}+3/2\,\sin
 \left( x \right)  \right) \cos \left( x \right) +3/8\,x
+C$$
and
$$\int\sin(x)^6dx=-1/6\, \left(  \left( \sin \left( x \right)  \right) ^{5}+5/4\,
 \left( \sin \left( x \right)  \right) ^{3}+{\frac {15\,\sin \left( x
 \right) }{8}} \right) \cos \left( x \right) +{\frac {5\,x}{16}}
+C$$
A: Expanding on @Dr.SonnhardGraubner's answer, since $\sin^2x=\frac{1-\cos 2x}{2}$ we have $$\sin^4x=\frac{1-2\cos 2x+\cos^22x}{4}=\frac{3-4\cos 2x+\cos4x}{8}$$ while $$\sin^6x=\frac{1-3\cos 2x+3\cos^22x-\cos^32x}{8}=\frac{10-15\cos 2x+6\cos4x-\cos6x}{32}.$$Thus $$\int(\sin^4x-\sin^6x)dx\\=\int\frac{2-\cos2x-2\cos4x+\cos6x}{32}dx=\frac{12x-3\sin2x-3\sin4x+\sin6x}{192}+C.$$
A: The integral is elementary, that is, expressible as a combination of finitely many elementary functions. Thus, this doesn't say anything about how arduous or difficult the process of transforming the integral into recognizable form may be.
That said, there are other ways to proceed. One of them is described as follows, where we simplify the integrand.
So note that we have $\cos^2x\sin^2x=\frac14\sin^22x=\frac18(2\sin^22x)=\frac18(1-\cos 4x).$ It follows that $$\cos^2x\sin^4x=\frac18(1-\cos 4x)\frac12(1-\cos 2x)=\frac{1}{16}(1-\cos 2x-\cos 4x+\cos2x\cos 4x)=\frac{1}{16}[1-\cos 2x-\cos 4x+\frac12(\cos 6x+\cos 2x)],$$ from where you can continue.
A: Alternative solution:
$$\int \cos^2x\sin^4x\mathrm{d}x=\\
\int(\frac{1+\cos (2x)}{2})(\frac{1+\cos (2x)}{2})^2\mathrm{d}x=\\\frac{1}{8}
\int (1+\cos (2x))(1-2\cos (2x)+\cos^2(2x))\mathrm{d}x=
\\\frac{1}{8}
\int (1+\cos (2x))(1-2\cos (2x)+\frac{1+\cos(4x)}{2})\mathrm{d}x=$$
A: For trigonometric monomials, the standard method is linearisation when both exponents are even, substitution when an exponent is odd ($u=\sin x$ or $u=\cos x$, depending on which exponent is odd).
You can linearise either with trigonometry formulæ or via the complex exponential. I'll show both:


*

*with trigonometry: use repeatedly the linearisation formulæ
$$\cos^2x=\tfrac12(1+\cos 2x),\qquad\sin^2x=\tfrac12(1-\cos 2x).$$
So
\begin{align}
\cos^2x\sin^4x&=\tfrac12(1+\cos 2x)\tfrac14(1-\cos 2x)^2=\tfrac 18(1+\cos 2x)(1-2\cos 2x+\cos^22x)\\
&=\tfrac 18(1+\cos 2x)\bigl(\tfrac32-2\cos 2x+\tfrac12\cos4x)\\
&=\tfrac1{16}(3-4\cos 2x+\cos 4x+3\cos 2x-4\cos^2 2x+\underbrace{\cos2x\cos 4x}_{\frac12(\cos 6x+\cos 2x)}) \\ 
&=\tfrac1{32}(2-\cos 2x-2\cos 4x+\cos 6x).
\end{align}

*with the complex exponential: to have lighter computations, set $u=\mathrm e^{ix}$. We'll use the definitipn of sine and cosine in terms of $u$:
$$\sin x=\tfrac1{2i}(u-\bar u),\qquad \cos x =\tfrac 12(u+\bar u),$$
so that, taking into accoount that $u\bar u=1$,
\begin{align}
\cos^2x\sin^4x &=\tfrac14(u+\bar u)^2\cdot\tfrac1{16}(u-\bar u)^4=\tfrac1{64}(u^2+\bar u^2+2)(u^4-4^2+6-4\bar u^2+\bar u^4) \\
&=
\end{align}
A: Here is to integrate economically,
$$\cos^2x\sin^4 x = \frac 18 \sin^2 2x (1-\cos 2x)= \frac {1}{16}-\frac {1}{16}\cos 4x-\frac 18 \sin^22x\cos 2x$$
Thus,
$$\int \cos^2x\sin^4x\mathrm{d}x = \frac {x}{16}-\frac {1}{64}\sin4x-\frac{1}{48}\sin^32x +C$$
A: If you insist on the tangent substitution, I think you will be mildly disappointed... Taking $t=\tan x$ makes it so
$$\begin{cases}
\sin x=\frac t{\sqrt{1+t^2}}\\[1ex]
\cos x=\frac1{\sqrt{1+t^2}}\\[1ex]
\mathrm dt=\sec^2x\,\mathrm dx\implies\mathrm dx=\frac{\mathrm dt}{1+t^2}
\end{cases}$$
The integral becomes
$$\int\cos^2x\sin^4x\,\mathrm dx=\int\frac1{1+t^2}\frac{t^4}{(1+t^2)^2}\frac{\mathrm dt}{1+t^2}=\int\frac{t^4}{(1+t^2)^4}\,\mathrm dt$$
Decompose the integrand into partial fractions:
$$\frac{t^4}{(1+t^2)^4}=\frac1{(1+t^2)^2}-\frac2{(1+t^2)^3}+\frac1{(1+t^2)^4}$$
But then the obvious means of tackling these integrals is to undo the original one:
$$\int\frac{\mathrm dt}{(1+t^2)^2}=\int\frac{\sec^2x}{(1+\tan^2x)^2}\,\mathrm dx=\int\cos^2x\,\mathrm dx$$
$$-\int\frac2{(1+t^2)^3}\,\mathrm dt=-2\int\frac{\sec^2x}{(1+\tan^2x)^3}\,\mathrm dx=-2\int\cos^4x\,\mathrm dx$$
$$\int\frac{\mathrm dt}{(1+t^2)^4}=\int\frac{\sec^2x}{(1+\tan^2x)^4}\,\mathrm dx=\int\cos^6x\,\mathrm dx$$
so all that work to confirm what other answers have suggested: write the integrand in terms of powers of either $\cos x$ or $\sin x$, then use the well-known power reduction identities.
