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Integrate $$\int \frac{1}{\sin^4(x)\cos^4(x)}dx$$.

So I know that for this one we have to use a identity or u substitution, integration by parts in probably not going to help....can someone please point out what should I do to evaluate this integral? Thanks!

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The Maple command $$Student[Calculus1]:-IntTutor(1/(sin(x)^4*cos(x)^4), x) $$ finds it step by step with explanations. See here for info. –  user64494 Sep 27 '13 at 17:23

3 Answers 3

up vote 2 down vote accepted

\begin{align*} \int \frac{16}{16 \sin^4(x)\cos^4(x)}dx &= 16 \int \frac{1}{(2 \sin(x)\cos(x))^4} dx \\ &= 16 \int \csc^4(2x)dx \\ & = 16 \int \csc^2(2x) \csc^2(2x)dx\\ &= 16 \int (1 + \cot^2(2x)) \csc^2(2x) dx \\ &= 16 \int \csc^2(2x)dx + 16 \int \cot^2(2x)\csc^2(2x)dx \\ &= \; \;\dots \\ &= \; \; \dots \\ \end{align*}

For the last let, $\cot(2x) = u$

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Hint: Substitute $t=\tan x$.

Then $\frac{dx}{\sin^4 x \cos^4 x} = \frac{(1+t^2)^3}{t^4}dt$.

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All integrals of rational functions of $sin(x)$ and $cos(x)$ can be solved by rationally parameterizing the unit circle ( see http://mathnow.wordpress.com/2009/11/06/a-rational-parameterization-of-the-unit-circle/ for example). This will convert the integral into a rational integral, which are all solvable by partial fraction decomposition. Try this out for the infamous $\int sec^3(x)dx$ for instance.

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