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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!

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    $\begingroup$ There is a general method when the expression can be written as $$\int P(\cos x,\sin x)\,dx,$$ namely, the tangent half-angle substitution. Substitituing $t=\tan(x/2)$ you get $\cos x = \frac{1-t^2}{1+t^2}$ and $\sin x = \frac{2t}{1+t^2}$ and $dx=\frac{2\,dt}{1+t^2}.$ In your original formula, you get: $$\frac{1}{2}\int \frac{(1+t^2)^5}{t^2(1-t^2)^4}\,dt$$ which you can then solve by partial fractions. $\endgroup$ Commented Oct 29, 2019 at 17:21
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    $\begingroup$ Your more general formula is $$\frac{1}{2^{2k-1}}\int \frac{(1+t^2)^{2(p+k)-1}}{t^{2k}(1-t^2)^{2p}}$$ $\endgroup$ Commented Oct 29, 2019 at 17:24
  • $\begingroup$ @ThomasAndrews Thank you, that is very helpful! $\endgroup$ Commented Oct 29, 2019 at 17:50

4 Answers 4

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

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

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    $\begingroup$ The more general equation is $$\int \frac{1}{\sin^{2k} x\cos^{2p}x}\,dx =\int \frac{(1+t^2)^{k+p-1}}{t^{2k}}\,dt$$ where $t=\tan x.$ $\endgroup$ Commented Oct 29, 2019 at 17:34
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$$ \int {\sin^2x + \cos^2x\over \sin^2x\cos^4x}dx=\int {dx\over \cos^4x}+\int {dx\over \sin^2x\cos^2x}\\ =\int {(\tan^2x+1)\,d\tan x}+4\int {dx\over \sin^22x}\\ =\frac13\tan^3x+\tan x-2\cot2x.$$

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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.

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Another method is to substitute $t=\cot x$:

$$\begin{align}\int\frac{\mathrm dx}{\sin^2x\cos^4x}&=-\int\frac{(t^2+1)^2}{t^4}\mathrm dt\\&=-\int1+\frac2{t^2}+\frac1{t^4}\mathrm dt\\&=-t+\frac2{t}+\frac1{3t^3}+C\\&=-\cot x+2\tan x+\frac{\tan^3x}3+C\end{align}$$

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