Evaluate $\int \frac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}dx$ $$\int \dfrac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}$$
My multiple attempts are as follows:-
Attempt $1$:
$$\int \dfrac{\sqrt{1-2\sin^2x\cos^2x}\cdot\sin x}{\sin^4x\cos x}dx$$
$$\cos x=t$$
$$-\sin x=\dfrac{dt}{dx}$$
$$\int -\dfrac{\sqrt{1-2t^2(1-t^2)}}{t(1-t^2)^2}dt$$
$$\int -\dfrac{\sqrt{2t^4+1-2t^2}}{t(1+t^4-2t^2)}dt$$
$$\int -\dfrac{\sqrt{2t^2+\dfrac{1}{t^2}-2}}{1+t^4-2t^2}dt$$
$$\int -\dfrac{\sqrt{2t^2+\dfrac{1}{t^2}-2}}{\left(\dfrac{1}{t^2}+t^2-2\right)t^2}dt$$
$$\int -\dfrac{\sqrt{\dfrac{2}{t^2}+\dfrac{1}{t^6}-\dfrac{2}{t^4}}}{\left(\dfrac{1}{t^2}+t^2-2\right)}dt$$
Not finding the way to proceed from here.
Attempt $2$:
$$\int \dfrac{\sqrt{\tan^4x+1}\cos x}{\sin^3 x}dx$$
$$\int \dfrac{\sqrt{\tan^4x+1}}{\tan^3 x\cos^2x}dx$$
$$\tan x=t$$
$$\int \dfrac{\sqrt{t^4+1}}{t^3}dt$$
$$t=\sqrt{\tan\theta}$$
$$\dfrac{dt}{d\theta}=\dfrac{1}{2\sqrt{\tan\theta}}\sec^2\theta$$
$$\int \dfrac{\sec\theta\sec^2\theta}{2\tan^2\theta}d\theta$$
$$\int \dfrac{\sec\theta}{2\sin^2\theta}d\theta$$
$$\int \dfrac{\cos\theta}{2\sin^2\theta\cos^2\theta}d\theta$$
$$y=\sin\theta$$
$$\int \dfrac{dy}{2y^2(1-y^2)}$$
$$-\dfrac{1}{2}\cdot\int\dfrac{y^2-(y^2-1)}{y^2(y^2-1)}dy$$
$$-\dfrac{1}{2}\cdot\left(\dfrac{1}{2}\cdot\ln\left|\dfrac{y-1}{y+1}\right|+\dfrac{1}{y}\right)+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{\sin\theta-1}{\sin\theta+1}\right|-\dfrac{1}{2\sin\theta}+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{t^2-\sqrt{1+t^4}}{t^2+\sqrt{1+t^4}}\right|-\dfrac{\sqrt{1+t^4}}{2t^2}+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{\tan^2x-\sqrt{1+\tan^4x}}{\tan^2x+\sqrt{1+\tan^4x}}\right|-\dfrac{\sqrt{1+\tan^4x}}{2\tan^2x}+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{1-\sqrt{\cot^4x+1}}{1+\sqrt{\cot^4x+1}}\right|-\dfrac{\sqrt{1+\cot^4x}}{2}+C$$
But real answer is $-\dfrac{\mathrm{cosec}x}{2}-\dfrac{1}{4}\cdot\ln\left|\dfrac{\mathrm{cosec}x-1}{\mathrm{cosec}x+1}\right|+C$
What am I missing here?
 A: Claim. The "real answer" is incorrect.
Proof. Let $\csc(x)=\frac1{\sin(x)}$ and $\sec(x)=\frac1{\cos(x)}$. Then, for the given solution at points where $\frac{\csc(x)-1}{\csc(x)+1}>0$,  $$\frac{\mathrm d}{\mathrm dx}\bbox[10px,#ffd]{\frac{1}{4} \big(-2\csc (x)-\log (\csc (x)-1)+\log (\csc (x)+1)\big)}=\frac{\csc(x)}{4} \left(2 \cot (x)+\underbrace{\frac{\cot(x)}{\csc (x)-1}-\frac{\cot
   (x)}{\csc (x)+1}}_{\large =2\tan(x)}\right)=\frac{\csc(x)}2(\tan(x)+\cot(x))=\csc(x)\csc(2x)$$
which does not equal your original function. $\square$
Claim. Your solution is correct.
(Constructive) proof. After $$\int \frac{\sqrt{t^4+1}}{t^3}\,\mathrm dt$$ we can continue with the substitution $u=t^{-4}$ and $\,\mathrm du=-4t^{-5}\,\mathrm dt$ to obtain $$-\frac14\int \frac{\sqrt{u+1}}u\,\mathrm du=-\frac12\int\frac{u+1}{2u\sqrt{u+1}}\,\mathrm du\overset{s=\sqrt{u+1}}=-\frac12\int\frac{s^2}{s^2-1}\,\mathrm ds=-\frac s2+\frac14\ln\left|\frac{s+1}{s-1}\right|+C$$
Since $s=\sqrt{1+t^{-4}}=\sqrt{1+\cot^4(x)}$, we get exactly your expression. $\square$
Remark. The integral can also be written as $$\frac{1}{2} \operatorname{arcsinh}\left(\tan ^2(x)\right)-\frac{1}{2} \sqrt{\tan ^4(x)+1} \cot
   ^2(x)$$
A: Continue  the second approach below with integration by parts
$$I=\int \dfrac{\sqrt{t^4+1}}{t^3}d
=-\frac12 \frac{1+t^4}{t^2} +\int \frac{tdt}{\sqrt{1+t^4}}dt$$
where 
$$\int \frac{tdt}{\sqrt{1+t^4}}=\frac12 \int \frac{d(t^2)}{\sqrt{1+(t^2)^2}}=\frac12 \sinh^{-1}t^2
$$
Thus,
$$I =- \frac{1+t^4}{2t^2} +\frac12 \sinh^{-1}t^2+C$$
