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

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?

• that's not right. $\sin(2x) = 2\sin(x)\cos(x)$ – mathreadler May 21 '17 at 10:47
• $t = \sin (x) , dt = \cos (x) dx$ could maybe give a hint – mathreadler May 21 '17 at 10:49
• You do not want to "find" this integral because you have it. You want to "find an expression without integral sign for..." – Jean Marie May 21 '17 at 10:55
• the solution is given by $$\frac{1}{4} \left(\frac{\sin ^2(x)}{\sin ^4(x)+1}+\tan ^{-1}\left(\sin ^2(x)\right)\right)$$ – Dr. Sonnhard Graubner May 21 '17 at 10:56

[![ 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 .

• can you please explain how you did an integration by parts? Which part of this function $\frac{-u^{2}}{(1+u^2)^2}$ did you derivate? – Green Banana May 21 '17 at 12:18
• sure , i chose to derivate u , so that i can integrate -u/((1+u^2)^2) – Youssef Khiari May 21 '17 at 16:18
• if i'm wrong or if there is a better way to solve please let me know . – Youssef Khiari May 21 '17 at 16:19

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

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

• yes this is good advice, also $t=\sin(x)$ could probably work. – mathreadler May 21 '17 at 10:51

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