Integral question using trig identities

Question

$$\int{\theta \tan^3{(\theta^2)}\sec^4{(\theta^2)}d\theta}$$

I thought of the method of splitting up the $\tan^3 \theta^2$ to $\tan^2 \theta^2$ and $\tan\theta^2$.

And then using trig identity $1+\tan^2\theta=\sec^2\theta$ to express the whole integral in terms of $\sec\theta$.

And then try to manipulate it into $\sec\theta \tan\theta$ to integrate the expression in terms of $d(\sec\theta)$.

That was my plan, but I got stuck right after splitting up the $\tan^3\theta$ and converting it to $\tan\theta^2 (\sec^2\theta^2 -1)$, because I didn't know what to do with the $\theta$ at the very front.

Answer 1

$$=\int \theta\tan^3(\theta) ^2\sec^4(\theta)^2 d\theta$$ Put $\theta^2 =x$ Hence we get $$=\frac 12\int \tan^3 x\sec^4 x dx$$ Put $u=\tan x$

$$=\frac 12\int u^3(1+u^2)du=\frac 12\int u^3+\frac 12\int u^5 du=\frac {\tan^4 (\theta)^2}{8}+\frac {\tan^6(\theta)^2}{12}+C$$

Answer 2

Substitute $\varphi$ = $\theta^2$. You would get $d\varphi=2\theta d\theta$.

$$\int{\theta \tan^3{(\theta^2)}\sec^4{(\theta^2)}d\theta} = \frac{1}{2}\int{\tan^3{(\varphi)}\sec^4{(\varphi)}d\theta}$$

This should get the lone $\theta$ which bothers the integration. Now you got the integral in terms of just trignometric functions which you can continue from here.