Integrating trigonometric functions. I need a little help with solving the following two integrals. I am not able to approach these two problems correctly. Little hints will suffice. Thanks for your help!


*

*$$\int\frac{\tan^2x\sec^2x}{1+ \tan^6x}\,dx$$

*$$\int (\sin x \cos x)^{1/3}\,dx$$
 A: *

*For the first integral, use the fact that $\tan'x=\sec^2x$.

*The second one cannot be expressed in terms of elementary functions. However, its definite counterpart can be written in terms of the beta function, using Wallis' integrals: $$\int_0^\tfrac\pi2\sin^{2n-1}(x)~\cos^{2m-1}(x)~dx=\dfrac{B(m,n)}2$$ In our case, $m=n=2/3$.
A: $1.$
$\int\dfrac{\tan^2x\sec^2x}{1+\tan^6x}dx$
$=\int\dfrac{\tan^2x}{1+\tan^6x}d(\tan x)$
$=\int\dfrac{1}{3(1+\tan^6x)}d(\tan^3x)$
$=\dfrac{\tan^{-1}\tan^3x}{3}+C$
$2.$
$\int(\sin x\cos x)^\frac{1}{3}dx=\int\dfrac{\sin^\frac{1}{3}2x}{2^\frac{1}{3}}dx$
Let $u=\sin^\frac{1}{3}2x$ ,
Then $x=\dfrac{\sin^{-1}u^3}{2}$
$dx=\dfrac{3u^2}{2\sqrt{1-u^6}}du$
$\therefore\int\dfrac{\sin^\frac{1}{3}2x}{2^\frac{1}{3}}dx$
$=\int\dfrac{3u^3}{2^\frac{4}{3}\sqrt{1-u^6}}du$
$=\int\dfrac{3u^3}{2^\frac{4}{3}}\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{6n}}{4^n(n!)^2}du$
$=\int\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{6n+3}}{2^{2n+\frac{4}{3}}(n!)^2}du$
$=\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{6n+4}}{2^{2n+\frac{4}{3}}(n!)^2(6n+4)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{6n+4}}{2^{2n+\frac{7}{3}}(n!)^2(3n+2)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{3(2n)!\sin^{2n+\frac{4}{3}}2x}{2^{2n+\frac{7}{3}}(n!)^2(3n+2)}+C$
A: Question 1:
Let $u=\tan^3x$.
\begin{align*}
&\int\frac{\tan^2x\sec^2x}{1+\tan^6x}\,dx\\
=&\int\frac{\tan^2x\sec^2x}{1+u^2}\frac{du}{3\tan^2x\sec^2x}\\
=&\frac13\int\frac{du}{1+u^2}\,du\\
=&\frac13\arctan u+C\\
=&\frac13\arctan \tan^3x+C
\end{align*}
Question 2:
Does not look like it has a closed form. Are there any limits?
