3
$\begingroup$

How can I integrate $$\int \frac{1}{\sin^4(x)\cos^4(x)}\,\mathrm dx.$$

So I know that for this one we have to use a trigonometric identity or a substitution. Integration by parts is probably not going to help. Can someone please point out what should I do to evaluate this integral?

Thanks!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user64494 Sep 27 '13 at 17:23
2
$\begingroup$

\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, $\cot(2x) = u$

$\endgroup$
2
$\begingroup$

Hint: Substitute $t=\tan x$.

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.