Integrate $\int\sin^2(x)\cos^4(x)dx$

Question

As in the title: how does one integrate $$ \int\sin^2(x)\cos^4(x)dx? $$

Any hints? Integration by parts doesn't seem to be a reasonable technique here.

Answer 1

In idea for you to develop:

$$\sin x\cos x=\frac12\sin 2x\implies \sin^2x\cos^4x=\left(\frac12\sin2x\right)^2\cos^2x=$$

$$=\frac14\sin^22x\left(\frac12\left(\cos2x+1\right)\right)=\frac18\cos2x\sin^22x+\frac18\sin^22x$$

Observe now that

$$\left(\sin2x\right)'=2\cos 2x,\,\text{ and}\;\;\int\sin^22x\,dx=\frac12\int\sin^2u\,du=\frac14\left(u-\sin u\cos u\right)=...$$

Answer 2

Or you could separate it into two integrals right from the beginning:

$$ \int \sin^2 x \cos^4 x \ dx = \int \cos^4 x \ dx - \int \cos^6 x \ dx. $$

Then, we have

$$ \cos^4 x = \frac{3 + \cos 4x + 4\cos 2x}{8} $$

and

$$ \cos^6 x = \frac{10 + \cos 6x + 6 \cos 4x + 15 \cos 2x}{32}. $$

On completing the integration, the answer should be:

$$ \int \sin^2 x \cos^4 x \ dx = \frac{x}{16} + \frac{\sin 2x}{64} - \frac{\sin 4x}{64} - \frac{\sin 6x}{192}+C. $$

Answer 3

Setting $u = \sin(x), \; dv = \cos^4(x)\sin(x) \, dx$, we get

$$\int \sin^2(x) \cos^4(x) \, dx = -\sin(x)\frac{\cos^5(x)}{5} + \frac{1}{5}\int \cos^6(x) \, dx$$

which certainly doesn't look any better, but we can take two cosines from the six and convert them into sines to get

$$\int (1-\sin^2(x)) \cos^4(x) \, dx = \int \cos^4(x) \,dx - \int\sin^2(x) \cos^4(x) \, dx$$

but that's the same as our original integral, so we can move it over to the LHS to get

$$\frac{6}{5} \int \sin^2(x) \cos^4(x) \, dx = -\sin(x)\frac{\cos^5(x)}{5} + \frac{1}{5}\int \cos^4(x) \, dx$$

so that we now have a reduced problem. In fact, you could even do a similar trick to evaluate the remaining integral.

Answer 4

This is not as neat as the answer by DonAntonio, but it works: \begin{align} \int\sin^2(2x)\cos^4(x)dx&=\int\frac{1-\cos(2x)}{2}\frac{3+4\cos(2x)+\cos(4x)}{8}dx \end{align}