<s>[INST] Evaluate $\int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}}dx$ $$I = \int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}}$$ 
Thought for a while but cannot seem to find how do i procced with this integral. 
Another variant of this problem is $$I = \int \frac{\sin^{3}x+\cos^{3}x}{\sin^{2}{x}\cos^{2}{x}}$$  where you just seperate the terms in numerator and it simplifies to $$I=\int \sec{x}\tan{x}dx+\int \csc{x}\cot{x}dx$$ which gives $$\sec{x}-\csc{x}+C$$ on integrating. 
But in this case it doesn't simplify into something straightforward. 
 [/INST] Notice that
\begin{eqnarray}
\dfrac{\sin^3x+\cos^3x}{\sin x\cos x}&=&
\dfrac{\sin^3x}{\sin x\cos x}+
\dfrac{\cos^3x}{\sin x\cos x}\\
&=&\dfrac{\sin^2x}{\cos x}+\dfrac{\cos^2x}{\sin x}\\
&=&\dfrac{1-\cos^2x}{\cos x}+\dfrac{1-\sin^2x}{\sin x}\\
&=&\dfrac{1}{\cos x}-\dfrac{\cos^2x}{\cos x}+\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}\\
&=&\dfrac{1}{\cos x}+\dfrac{1}{\sin x}-\cos x-\sin x
\end{eqnarray}
To compute the two integrals
$$
\int\dfrac{1}{\cos x}\,dx \text{  and } \int\dfrac{1}{\sin x}\,dx
$$
we use the substitution
$$
t=\tan\dfrac{x}{2} \text{ or equivalently } x=2\tan^{-1}t.
$$
Since
\begin{eqnarray}
\cos x&=&\dfrac{\cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}}=\dfrac{1-t^2}{1+t^2}\\
\sin x&=&\dfrac{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}}=\dfrac{2t}{1+t^2}\\
dx&=&\dfrac{2}{1+t^2}\,dt
\end{eqnarray}
we have
\begin{eqnarray}
\int\dfrac{1}{\cos x}\,dx&=&\int\dfrac{1+t^2}{1-t^2}\cdot\dfrac{2}{1+t^2}\,dt=\int\dfrac{2}{1-t^2}\,dt=\int\left(\dfrac{1}{1-t}+\dfrac{1}{1+t}\right)\,dt=\ln\left|\dfrac{1+t}{1-t}\right|+c_1\\
\int\dfrac{1}{\sin x}\,dx&=&\int\dfrac{1+t^2}{2t}\cdot\dfrac{2}{1+t^2}\,dt=\int\dfrac{1}{t}\,dt=\ln|t|+c_2
\end{eqnarray}
Hence
\begin{eqnarray}
\int\dfrac{\sin^3x+\cos^3x}{\sin x\cos x}\,dx&=&\int\left(\dfrac{1}{\cos x}+\dfrac{1}{\sin x}-\cos x-\sin x\right)\,dx\\
&=&\ln\left|\dfrac{1+\tan\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}\right|+\ln\left|\tan\dfrac{x}{2}\right|-\sin x+\cos x+c
\end{eqnarray}
</s>