Solve this integral $\int \frac{dx}{\sin^6x \cdot \cos^6x}$ I was solving this integral manually
$$\int \frac{dx}{\sin^6x \cdot \cos^6x}$$
It's not easy but it wasn't that hard too having all good formulas/recurrences in front of me and not rushing my calculations.
So I got this answer by hand:
$$(1/5)  \cdot  \sin(x)^{-5}  \cdot  \cos(x)^{-5} +  \\  (2/3)  \cdot  \sin(x)^{-5}  \cdot  \cos(x)^{-3} +  \\    (16/3)  \cdot  \sin(x)^{-5}  \cdot  \cos(x)^{-1}  \\  -   (32/5)  \cdot  \sin(x)^{-5}  \cdot  \cos(x) \\ - (32 \cdot 4/15)  \cdot  \sin(x)^{-3}  \cdot  \cos(x) \\ -   (32 \cdot 8/15)  \cdot  \sin(x)^{-1}  \cdot  \cos(x)     $$
Now... my problem is Wolfram Alpha does not even want to compute the derivative of this. Is it too long for the free WA version? I then tried using SymPy Live but that does not help either. It is not able to simplify the derivative to $\sin^6x \cdot \cos^6x$, it seems.
So I finally found this free derivative calculator (not sure what it's based on),
and it seems to be saying my answer is OK.
www.derivative-calculator.net
Is my answer OK indeed?
What is the best free online tool I can use for this task?
 A: Probably the fastest way to make sure that your answer is correct is to integrate it in a faster way and check if the difference between the two answers is constant.
Note here that the integral is pretty easy:
$$\frac{1}{\sin^6x \cdot \cos^6x}=\frac{2^6}{ \sin^6(2x)} =2^6 \csc^6(2x)= 64 \csc^2(2x) \left( 1+ \cot^2(2x) \right)^2 $$
which is easy to integrate.
A: The answer is $$\boxed{\int \frac{1}{\sin^{6}(x)\cos^{6}(x)}dx=-\frac{1}{30}\left( 10\cos(2x)-5\cos(6x)+\cos(10x)\right)\csc^{5}(x)\sec^{5}(x)+C}.$$
where $C\in \mathbb{R}$.
Note that, $$I=\int \frac{1}{\sin^{6}(x)\cos^{6}(x)}dx$$
It's easy to see that $$I=\int \csc^{6}(x)\sec^{6}(x)dx$$
by reduction formula, we have that $$I=-\frac{1}{5}\csc^{5}(x)\sec^{5}(x)+2\int\csc^{4}(x)\sec^{6}(x)dx$$
so, we have other applications of reduction formula $$I=-\frac{2}{3}\csc^{3}(x)\sec^{5}(x)-\frac{1}{5}\csc^{5}(x)\sec^{5}(x)+\frac{16}{3}\int \csc^{2}(x)\sec^{6}(x)dx$$
Using $\sec^{2}(x)=\tan^{2}(x)+1$ and $\csc^{2}(x)=\cot^{2}(x)+1$, we have $$I=-\frac{2}{3}\csc^{3}(x)\sec^{5}(x)-\frac{1}{5}\csc^{5}(x)\sec^{5}(x)+\frac{16}{3}\int (\tan^{2}(x)+1)^{3}(\cot^{2}(x)+1)dx$$
Finally, let $u=\tan(x)$, you have $$I=-\frac{1}{30}\left( 10\cos(2x)-5\cos(6x)+\cos(10x)\right)\csc^{5}(x)\sec^{5}(x)+C$$
and you can see that $$\frac{d}{dx}\left( -\frac{1}{30}\left( 10\cos(2x)-5\cos(6x)+\cos(10x)\right)\csc^{5}(x)\sec^{5}(x)+C\right)=\csc^{6}(x)\sec^{6}(x).$$
A: Since your result has been confirmed.
It would have been very fast to use $x=\tan^{-1}(t)$ which makes
$$I=\int \frac{dx}{\sin^6(x) \, \cos^6(x)}=\int\frac{\left(t^2+1\right)^5}{t^6}\,dt $$ that is to say
$$I=\frac{t^5}{5}-\frac{1}{5 t^5}+\frac{5 t^3}{3}-\frac{5}{3 t^3}+10 t-\frac{10}{t}+C$$ Back to $x$
$$I=\frac{\tan ^5(x)}{5}+\frac{5 \tan ^3(x)}{3}+10 \tan (x)-\frac{1}{5} \cot
   ^5(x)-\frac{5 \cot ^3(x)}{3}-10 \cot (x)+C$$ Simplifying
$$I=-\frac{1}{30} (10 \cos (2 x)-5 \cos (6 x)+\cos (10 x)) \csc ^5(x) \sec ^5(x)+C$$
