A simpler way to evaluate $\int {dx\over \sin^2x\cos^4x}$

Question

Evaluate: $$ \int {dx\over \sin^2x\cos^4x} $$

I've started by using identities: $$ {1\over \sin^2x} = 1+\cot^2x\\ {1\over \cos^2x} = 1+\tan^2x $$

So the integral becomes: $$ \begin{align} I &= \int (1+\cot^2x)(1+\tan^2x)^2dx \\ &=\int (1+\cot^2x)(1+2\tan^2x + \tan^4x)dx \\ &=\int (1 + 2\tan^2x + \tan^4x+\cot^2x + 2 + 2\tan^2x)dx \\ &=\int (3 + 4\tan^2x+\tan^4x+\cot^2x)dx \end{align} $$

I know the integral of $\tan^2x$ and $\cot^2x$, while for $\tan^nx$ one could use a reduction formula. But I find this approach too complicated and seek for a simpler one. Another approach I've tried out is expanding $1$ in the nominator to: $$ \int {\sin^2x + \cos^2x\over \sin^2x\cos^4x}dx $$

But that was also clumsy.

Is there another approach simpler than the one suggested above? Could this approach be generalized for integrals of the form: $$ \int {dx\over \sin^{2k}x\cos^{2p}x} $$

where $k, p \in\Bbb N$?

Thank you!

Answer 1

$I = \displaystyle\int \dfrac{1}{\sin^2{x}\cos^4{x}} dx = \displaystyle\int \sec^2{x}\dfrac{\left(\tan^2{x}+1\right)^2}{\tan^2{x}} dx$. Now substitute $y = \tan{x}$.

$I= \displaystyle\int \dfrac{(y^2+1)^2}{y^2} dy = \frac{y^3}{3}+2y-\frac{1}{y} +C$

Answer 2

$$ \int {\sin^2x + \cos^2x\over \sin^2x\cos^4x}dx=\int {dx\over \cos^4x}+\int {dx\over \sin^2x\cos^2x}\\ =\int {(\tan^2x+1)\,d\tan x}+4\int {dx\over \sin^22x}\\ =\frac13\tan^3x+\tan x-2\cot2x.$$

Answer 3

Write $$I = \int \frac{dx}{\sin^2(x)\cos^4(x)} = \int \csc^2(x) \sec^4(x) dx$$ and use the trigonometric identities:

1. $\sec^2(x) - \tan^2(x) = 1$

2. $\csc^2(x) - \cot^2(x) = 1$

$$I = \int (\cot^2(x) + 1)(\tan^2(x) +1)^2 dx$$ Now use the substitution $u = \tan(x)$; from here the details should be straight-forward.