Integral of $\int \sin^2x\cos^4xdx$

Question

$$\int \sin^2x\cos^4xdx$$

I tried $$I = \int (1-\cos^2x)\cos^4xdx = \int \frac{\sec^2x-1}{\sec^6x}dx = \int \frac{\tan^2x}{\sec^6x}dx$$ Take $\tan x = t \implies \sec^2xdx = dt$ $$I = \int \frac{t^2}{(t^2+1)^4}dt$$

And I could not proceed further from here.

Answer 1

You can calculate it directly without substitution as follows:

$$\sin^2x\cos^4x=(\sin x\cos x)^2\cos^2x =\frac 14\sin^22x\cdot \frac 12(1+\cos 2x)$$ $$= \frac 18 \cdot\frac 12(1-\cos 4x) +\frac 18\sin^2 2x\cos 2x$$

Hence,

$$\int \sin^2x\cos^4x\; dx = \frac x{16}-\frac{\sin 4x}{64} + \frac{\sin^3 2x}{48}+c$$

Answer 2

Hint:

$$\dfrac{d(\sin x\cos^nx)}{dx}=\cos^{n+1}x-n\sin x\cos^{n-1}x\sin x$$

$$=\cos^{n+1}x-n\cos^{n-1}x(1-\cos^2x)$$

$$=(n+1)\cos^{n+1}x-n\cos^{n-1}x$$

Integrate both sides wrt $x,$ $$\implies(n+1)I_{n+1}=nI_n+\sin x\cos^nx+K$$ where $\displaystyle I_m=\int\cos^mx\ dx$

We need $$\int(\cos^4x-\cos^6x)dx$$

Answer 3

$$I=\int \sin^2 x \cos^4 x dx =\frac{1}{8} \int \sin^2 2x (1+\cos 2x) dx=\frac{1}{8}\int \sin ^2 2x dx+\frac{1}{8}\int (t^2/2) dt$$ Here $\sin 2x=t$ $$\implies I=\frac{1}{16}\int (1-\cos 4x) dx +\frac{(\sin 2x)^3}{48} =\frac{1}{16} x-\frac{1}{64} \sin 4x+\frac{(\sin 2x)^3}{48} $$

Answer 4

With even powers of the trig functions use the half-angle identities until you have a an odd power.

$\sin^2 x = \frac 12 (1-\cos 2x)\\ \cos^2 x = \frac 12 (1+\cos 2x)$

$(\sin^2 x)(\cos^4 x) = \frac {1}{8}(1-\cos 2x)(1+cos 2x)(1+\cos 2x)\\ \frac18(1-\cos^2 2x)(1+\cos 2x)\\ \frac 18 (1 +\cos 2x - \cos^2 2x - \cos^3 2x)\\ \frac 18 (1 +\cos 2x - \frac 12 (1+\cos 4x) - \cos 2x(1-\sin^2 2x))\\ \frac 1{16} (1 - \cos 4x - 2\sin^2 2x\cos 2x)$

And each of those terms is straight forward to integrate.

Alternatively, you can use some complex analysis and say:

$\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\ \cos x = \frac {e^{ix} + e^{-ix}}{2}$

$(\sin^2 x)(\cos^4 x) = \frac {(e^{2ix} -2 + e^{-2ix})(e^{4ix} + 4e^{2ix} + 6 + 4e^{-2ix} + e^{-4ix})}{-64}\\ \frac {e^{6ix} + 2e^{4ix} - e^{2ix} -4 - e^{-2ix} +2 e^{-4ix} + e^{-6ix}}{-64}\\ \frac {-\cos 6x - 2\cos 4x + \cos 2x + 2}{32}$

And again not to bad to integrate.