find an integral $\int \frac{\sin x \cos x}{(1+\sin ^{4}x)^{2}}dx$

Question

I need to find this integral $\int \frac{\sin x \cos x}{(1+\sin ^{4}x)^{2}}dx$.

I've tried to use these formulas: $\sin 2x = 2\sin x\cos x $
and $ \sin ^{2}x = \frac{1}{2}(1-\cos 2x)$.

And I came to this: $8\int \frac{\sin2x}{(4+(1-\cos 2x)^{2})^{2}} dx $.

Now I can use a substitution $1-\cos 2x = t$, $2\sin 2x dx = dt$.

And I have: $4\int \frac{1}{(4+t^{2})^{2}}dt$. But I don't know what to do with it. Integral calculator tells that I should apply the reduction formula here, but is there any way to solve this integral without it?

Answer 1

[![ try this ][1]][1]

[1]: https://i.stack.imgur.com/TmMHi.png try using u=sin(x)^2 ,it will be easier to integrate . you can use another way also .

Answer 2

Easier, use $sin(x)^2 = t$, $2\sin(x)\cos(x)dx = dt$. Try to resolve

$$ \int \frac{1}{2} \frac{dt}{(1+t^2)^2}$$

Answer 3

Put $\sin^2 x = \tan x$ and then the integral will boil down to $(\cos x)^2$ which can be solved by using $(\cos x )^2= (1+\cos x)/2$