# Evaluate $\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$

$$\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$$ $$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$ $$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$ $$\int \frac{(1- \cos2x)^2}{2.(1+\cos^2 2x)}{dx}$$ $$\frac{1}{2} \int \left[1-\frac{2 \cos2x}{1+\cos^22x}\right] dx$$

What should I do next ?

Please also tell me alternative way to do this .

• Now you can replace $\cos^2 2x$ by $1-\sin^2 2x$ in the denominator, and then let $u=\sin 2x$ to get $\frac{1}{2}\big[\int 1 dx-\int\frac{1}{2-u^2}du\big]$ Commented Jul 20, 2016 at 20:53
• This integral is somewhat similar: Indefinite integration problem $\int {\frac{1}{1+ \tan^4 x}}dx$. Found using Approach0. Commented Jul 21, 2018 at 7:29

HINT:

$$\cos^22x=1-\sin^22x$$

Set $\sin2x=y$

Otherwise,

$$\dfrac{\sin^4x}{\sin^4x+\cos^4x}=\dfrac1{1+\cot^4x}$$

Let $\cot^2x=u\implies dx=-\dfrac{du}{1+u^2}$

Method$\#1:\dfrac2{(1+u^4)(1+u^2)}=\dfrac{1+u^4+1-u^4}{(1+u^4)(1+u^2)}=?$

Method$\#2:$ Writing $u^2=y$

$$\dfrac1{(1+y^2)(1+y)}=\dfrac{Ay+B}{1+y^2}+\dfrac C{1+y}$$

$$\iff1=(Ay+B)(1+y)+C(1+y^2)=y^2(C+A)+y(B+A)+B+C$$

$C+A=0\iff C=-A, B+A=0\iff B=-A$ and $1=B+C=-2A$

$$\implies\dfrac1{(1+y^2)(1+y)}=\dfrac{-y+1}{2(1+y^2)}+\dfrac1{2(1+y)}$$

In either case, finally $\dfrac{1-u^2}{1+u^4}=\dfrac{\dfrac1{u^2}-1}{\dfrac1{u^2}+u^2}$

and $\displaystyle\int\left(\dfrac1{u^2}-1\right)=?$ and $\dfrac1{u^2}+u^2=\left(u+\dfrac1u\right)^2-2$

• How you did this $\cot^2x=u\implies dx=-\dfrac{du}{1+u^2}$ . Commented Jul 20, 2016 at 16:27
• $$-\csc^2x\ dx=du\iff du=?$$ Commented Jul 20, 2016 at 16:28

$$I=\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}=\int \frac{1-2\cos^2(x)+\cos^4(x)}{1-2\cos^2(x)+2\cos^4(x)}dx$$ $$=\int\frac{1}{2}-\frac 12\frac{-1+2\cos^2(x)}{1-2\cos^2(x)+2\cos^4(x)}dx$$ $$=\int\frac{1}{2}-\frac 12\frac{\cos(2x)}{1-2\cos^2(x)\sin^2(x)}dx$$ $\color{blue}{\text{This last step above is a few manipulations away from being the last step in your question.}}$

Let $u=\cos(x)\sin(x)$ then $du=\cos(2x)dx$ $$\boxed{I=\frac x2-\frac 12\int \frac 1{1-2u^2}du=\frac x2 -\frac12\frac{\tanh^{-1}(\sqrt2 u)}{\sqrt 2} +k_1}$$

Replace $u$ and you're good to go...

Hint: if we multiply by $\sin^{-6}\left(x\right)$ we get $$I=\int\frac{\sin^{4}\left(x\right)}{\sin^{4}\left(x\right)+\cos^{4}\left(x\right)}dx=\int\frac{\csc^{2}\left(x\right)}{\csc^{2}\left(x\right)+\cot^{4}\left(x\right)\csc^{2}\left(x\right)}dx$$ $$=\int\frac{\csc^{2}\left(x\right)}{\cot^{6}\left(x\right)+\cot^{4}\left(x\right)+\cot^{2}\left(x\right)+1}dx$$ and now taking $\cot\left(x\right)=u,\,-dx \csc^{2}\left(x\right)=du$ we have $$I=-\int\frac{1}{u^{6}+u^{4}+u^{2}+1}du=-\int\frac{1}{\left(u^{2}+1\right)\left(u^{4}+1\right)}du$$ $$=\int\frac{1}{\left(u^{2}+1\right)\left(u^{2}+\sqrt{2}x+1\right)\left(u^{2}-\sqrt{2}x+1\right)}du$$ can you take it from here?