Answer
The solution is $$\int\frac{\theta^3}{1+\sin\theta^4}d\theta=-\frac{1}{2(\tan\frac{\theta^4}{2}+1)}+c.$$
Work Step by Step
To solve the integral
$$\int\frac{\theta^3}{1+\sin\theta^4}d\theta$$
we will first use substitution $\theta^4=t$ which gives us $\theta^3d\theta=\frac{dt}{4}$. Substituting this we have:
$$\int\frac{\theta^3}{1+\sin\theta^4}d\theta=\frac{1}{4}\int\frac{dt}{1+\sin t}.$$
Now we will use new substitution $\tan\frac{t}{2}=x$ which gives us $dt=\frac{2}{1+x^2}dx,\sin t=\frac{2x}{1+x^2}.$ Putting this into the integral we have:
$$\frac{1}{4}\int\frac{dt}{1+\sin t}=\frac{1}{4}\int\frac{\frac{2}{1+x^2}}{1+\frac{2x}{1+x^2}}dx=\frac{1}{2}\int\frac{\frac{1}{1+x^2}}{\frac{1+x^2+2x}{1+x^2}}dx=\frac{1}{2}\int\frac{1}{1+x^2+2x}dx=\frac{1}{2}\int\frac{1}{(x+1)^2}dx.$$
Now we will use another substitution $x+1=a\Rightarrow dx=da.$ Putting this into the integral gives us:
$$\frac{1}{2}\int\frac{1}{(x+1)^2}dx=\frac{1}{2}\int\frac{1}{a^2}da=\frac{1}{2}\frac{-1}{a}+c$$
Now we have to express our solution in terms of $\theta$. To do this we will expres $a$ in terms of $x$, then $x$ in terms of $t$ and finally $t$ in terms of $\theta$:
$$\int\frac{\theta^3}{1+\sin\theta^4}d\theta=-\frac{1}{2a}+c=-\frac{1}{2(x+1)}+c=-\frac{1}{2(\tan\frac{t}{2}+1)}+c=
-\frac{1}{2(\tan\frac{\theta^4}{2}+1)}+c.$$
