Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I can approach the question? Since the integrand evaluates into an arctangent function I am assuming there is some trickery in the manipulation that can get me there. But I just can't seem to see it...
\begin{align} \int \frac{dx}{\sin^{4} x+\cos^{4} x} &= \int \frac{dx}{\frac14(1-\cos2x)^2+\frac14(1+\cos2x)^2}\\ &\hspace{-15mm}=\int \frac{2\ dx}{1+\cos^22x}=\int\frac{d(\tan 2x)}{2+\tan^22x} =\frac{1}{\sqrt{2}}\arctan\frac{\tan2x}{\sqrt{2}} \end{align}
Original post was to evaluate $$\int\frac{\,dx}{\sin^4(x)\cos^4(x)}$$ but I believe that the intended integral was $$\int\frac{\,dx}{\sin^4(x)+\cos^4(x)}$$ which I evaluate after the original request.
$$\begin{align}I=&\int\frac{\,dx}{\sin^4(x)\cos^4(x)}\\&=2^4\int\frac{\,dx}{\bigr[2\sin(x)\cos(x)\bigr]^4}\\&=2^4\int\frac{\,dx}{\sin^4(2x)}\\&=2^4\int\csc^4(2x)\,dx\\&=2^3\int \csc^4(u)\,du\end{align}$$
Using the reduction formula for cosecant, $$\int\csc^{m}(x)\,dx=-\frac{\csc^{m-1}(x)\cos(x)}{m-1}+\frac{m-2}{m-1}\int\csc^{m-2}(x)\,dx$$ or more concisely, $$J_m= -\frac{\csc^{m-1}(x)\cos(x)}{m-1}+\frac{m-2}{m-1}J_{m-2} $$ then we can obtain $$J_4=-\frac{\csc^{3}(u)\cos(u)}{3}+\frac{2}{3}J_2$$ where $J_2=-\cot(u)+C$
Let $$J_4=\int\csc^4{u}\,du$$ so that $$\begin{align}I&=2^3J_4\\&=2^3\bigg[-\frac{\csc^{3}(u)\cos(u)}{3}-\frac{2}{3}\cot(u)\bigg] +C\\&=-\frac{2^3}{3}\bigg[\csc^{3}(2x)\cos(2x)+2\cot(2x)\bigg] +C\\&=-\frac{2^3}{3}\cot(2x)\bigg[\csc^2(2x)+2\bigg]+C\end{align}$$
If you don't know the reduction formula off-hand or don't feel like deriving it, expand the integrand as $$\csc^4(u)=\csc^2(u)\csc^2(u)=\csc^2(u)(1+\cot^2(u))=\csc^2(u)+\csc^2(u)\cot^2(u)$$ which is fairly simple to integrate.
Now to evaluate $$I=\int\frac{\,dx}{\sin^4(x)+\cos^4(x)}$$
$$\begin{align}I&=\int\frac{\,dx}{\cos^4(x)-2\sin^2(x)\cos^2(x)+\sin^4(x)+2\sin^2(x)\cos^2(x)}\\&=\int\frac{\,dx}{[\cos^2(x)-\sin^2(x)]^2+2\sin^2(x)\cos^2(x)}\\&=\int\frac{\,dx}{\cos^2(2x)+\frac{\sin^2(2x)}{2}}\\&=\int\frac{\sec^2(2x)}{1+\frac{\tan^2(2x)}{2}}\,dx\\&=\int\frac{\sec^2(2x)}{1+\bigg(\frac{\tan(2x)}{\sqrt{2}}\bigg)^2}\,dx \end{align} $$
Let $u=\frac{\tan(2x)}{\sqrt{2}}$ and $\,du=\frac{2}{\sqrt{2}}\sec^2(2x)\,dx$. Then $$\begin{align}I&=\frac{\sqrt{2}}{2}\int\frac{\,du}{1+u^2}\\&=\frac{1}{\sqrt{2}}\arctan(u)+C\\&=\frac{1}{\sqrt{2}}\arctan\bigg(\frac{\tan(2x)}{\sqrt{2}}\bigg)+C \end{align} $$
$$\dfrac1{(\sin x\cos x)^4}=16\csc^42x=16\csc^22x(1+\cot^22x)$$
Set $\cot2x=y$
Alternatively
$$\dfrac1{(\sin x\cos x)^4}=\dfrac{(\cos^2x+\sin^2x)^2}{\cdots}=\csc^4x+\sec^4x+2\sec^2x\csc^2x$$
Now $\sec^2x\csc^2x=\dfrac{\cos^2x+\sin^2x}{\cos^2x\sin^2x}=?$
$$ \frac{1}{\sin^4x+\cos^4x}= \frac{1}{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}= \frac{1}{1-\frac{1}{2}(2\sin x\cos x)^2}=\\ \frac{2}{2-\sin^22x}= \frac{2}{2-\frac{1}{2}(1-\cos4x)}= \frac{4}{3+\cos4x}= $$ If we now set $$ t=\tan2x\qquad\implies\qquad dx=\frac{1}{2}\cdot\frac{1}{1+t^2}dt $$ we have $$ \int\frac{1}{\sin^4x+\cos^4x}dx= \int\frac{4}{3+\frac{1-t^2}{1+t^2}}\frac{1}{2(1+t^2)}dt= \int\frac{2}{3+3t^2+1-t^2}dt=\\ \int\frac{2}{2t^2+4}dt= \int\frac{1}{t^2+2}dt= \frac{1}{\sqrt{2}}\arctan\left(\frac{t}{\sqrt{2}}\right)+C= \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan(2x)}{\sqrt{2}}\right)+C= $$
OLD answer $$ \frac{1}{\sin^4x\cos^4x}=\frac{1}{(\sin x\cos x)^4}=16\frac{1}{\sin^42x}=16\frac{\cos^22x+\sin^22x}{\sin^42x}=\\=16\frac{\cot^22x+1}{\sin^22x}=-8\frac{d}{dx}\left(\frac{1}{3}\cot^32x+\cot2x\right) $$ Note that this is different from your supposed answer.
Hint: multiply nominator and denominator by $\sec^4(x)$, then use substitution $u=\tan (x)$.
Use $x = \arctan (t)$ substitution
So, $dx=\dfrac{dt}{1+t^2}$
$\sin^2x= \dfrac{\tan^2x}{1+\tan^2x} = \dfrac{t^2}{1+t^2}$
$\cos^2x= \dfrac{1}{1+\tan^2x} = \dfrac{1}{1+t^2}$
Can you take it from here?
Use the reduction formula, namely:
$$I=\int \csc^n(x)\sec^m(x)\,dx=-\frac{\csc^{m-1}(x)\sec^{n-1}(x)}{m-1}+\frac{m+n-2}{m-1}\int\csc^{-2+m}(x)\sec^n(x)\,dx.$$
Now set $n=4, m=4$ and you are left with
$$I=\tilde h(x)+2\int\sec^4(x)\csc^{2}(x)\,dx,$$
with $\tilde h(x)=-\frac{\csc^{3}(x)\sec^{3}(x)}{3}$.
The integral part of this result can be written as
$$\int\frac{(1+\tan^{2}(x))\sec^{2}(x)}{\tan^2(x)}\,dx$$
and it is solvable with the substitution $\tan(x)=t$. Indeed
$$\int\frac{(1+\tan^{2}(x))\sec^{2}(x)}{\tan^2(x)}\,dx=\int\frac{(1+t^2)^2}{t^2}=\frac{1}{3}\tan^3(x)+2\tan(x)-\cot(x)+c$$
and thus your final integral
$$I=\int \csc^n(x)\sec^m(x)\,dx=-\frac{\csc^{3}(x)\sec^{3}(x)}{3}+\frac{2}{3}\tan^3(x)+4\tan(x)-2\cot(x)+c.$$
$$\sin ^4(x)+\cos ^4(x)=\left(\sin ^2(x)+\cos ^2(x)\right)^2-2\sin^2x\cos^2x=1-\frac{1}{2}\sin^2(2x)=\\=1-\frac{1}{2}\cdot\frac{1-\cos(4x)}{2}=\frac{1}{4}\left(3+\cos(4x)\right)$$
$$\cos(4x)=\frac{1-\tan^2(2x)}{1+\tan^2(2x)}=\frac{1-t^2}{1+t^2}$$
where $\tan(2x)=t$, $x=\frac{1}{2}\arctan t$, $dx=\frac{dt}{2(1+t^2)}$
$$\int \frac{1}{\sin ^4(x)+\cos ^4(x)} \, dx=\int \frac{4}{3+\cos(4x)} \, dx=$$
$$=4\int \frac{1}{3+4\frac{1-t^2}{1+t^2}}\cdot \frac{dt}{2(1+t^2)}=\int \frac{dt}{2+t^2}$$
$t=u\sqrt{2}$, $dt=du\sqrt 2$
$$\int \frac{dt}{2+t^2}=\int \frac{du\sqrt 2}{2+2u^2}=\frac{\sqrt 2}{2}\int\frac{du}{1+u^2} = \frac{\sqrt 2}{2}\arctan u+C=\frac{\sqrt 2}{2}\arctan \frac{t}{\sqrt 2}+C=$$ $$=\frac{\sqrt 2}{2}\arctan \frac{\tan(2x)}{\sqrt 2}+C$$
$$ \begin{aligned} \int \frac{1}{\sin ^4 x+\cos ^4 x} d x = & \int \frac{1}{1-\frac{\sin ^2 2 x}{2}} d x \\ = & 2 \int \frac{1}{2-\sin ^2 2 x} d x \\ = & \int \frac{\sec ^2 2 x}{2 \sec ^2 2 x-\tan ^2 2 x} d(2 x) \\ = & \int \frac{d\left(\tan ^2 x\right)}{2\left(1+\tan ^2 2 x\right)-\tan ^2 2 x}\\=& \int \frac{d(\tan 2 x)}{2+\tan ^2 2 x} \\=&\int \frac{d(\tan 2 x)}{(\sqrt{2})^2+(\tan 2 x)^2}\\=&\frac{1}{\sqrt{2}}\arctan \left(\frac{\tan 2 x}{\sqrt{2}}\right)+C \end{aligned} $$
