Integrating $\int \frac{\sin^3x}{(\cos x)^\frac 4 3} dx$ 
Find: $\int \frac{\sin^3x}{(\cos x)^\frac 4 3} dx$

My attempt: 
Set $u=(\cos x)^\frac 4 3 $ so $du= \frac 4 3 (\cos x)^\frac 1 3 \sin x dx \Rightarrow dx= \frac 3 {4 (\cos x)^\frac 1 3 \sin x}du$
\begin{align}
\int \frac{\sin^3x}{(\cos x)^\frac 4 3} dx&=\frac 3 4 \int \frac{\sin ^2 x}{(\cos x)^\frac 5 3} du \\
&=\frac 3 4 \int \frac{1- \cos ^2 x}{(\cos x)^\frac 5 3} du\\
&=\frac 3 4 \int u^{-\frac 5 4}-u^\frac 1 4 du\\
&=-3(\cos^{-\frac 1 3} x +\frac 1 5 \cos ^{\frac 5 3 }x) +c
\end{align}
It's very close to the right answer, but not quite, what did I do wrong? Alternatively, is there a better way?
http://www.integral-calculator.com/#expr=%28sinx%29%5E3%2F%28cosx%29%5E%284%2F3%29

Let $u=\cos x$. Then $\mathrm du=-\sin x\mathrm dx$. Therefore
$$\begin{split}\int\frac{\sin^3x}{(\cos x)^{4/3}}\mathrm dx&=\int\frac{u^2-1}{u^{4/3}}\mathrm du\\&=\int \left(u^{2/3}-u^{-4/3}\right)\mathrm du\\&=\frac35u^{5/3}+3u^{-1/3}+c\\&=\frac35(\cos x)^{5/3}+3(\cos x)^{-1/3}
+c.\end{split}$$
