<s>[INST] $I=\int \frac{\cos^3(x)}{\sqrt{\sin^7(x)}}\,dx$ $$I=\int \frac{\cos^3(x)}{\sqrt{\sin^7(x)}}\,dx$$
I tried to write it as 
$$I=\int \sqrt{\frac{\cos^6(x)}{\sin^7(x)}}\,dx$$
And $$I=\int \sqrt{\frac{1}{\tan^6(x)\sin(x)}}\,dx$$ but it seems to go nowhere , how can I manipulate it so that its becomes solvable?
 [/INST] Here are the steps
$$\int \frac{\cos^3 x}{\sqrt{\sin^7 x}}dx= \int \frac{(1-\sin^2 x)(\cos x)}{\sqrt{\sin^7 x}}dx $$
Let $u=\sin x$, then $du=\cos x\ dx$. So now
$$ \int \frac{1-u^2}{\sqrt{u^7}}du = \int \frac{1}{\sqrt{u^7}}-\frac{u^2}{\sqrt{u^7}}du $$
$$= \int u^{-\frac72}du-\int u^{-\frac32}du $$
$$= \frac{u^{-\frac72+1}}{-\frac72+1}- \frac{u^{-\frac32+1}}{-\frac32+1}+C $$
$$= \frac{u^{-\frac52}}{-\frac52}- \frac{u^{-\frac12}}{-\frac12} +C $$
$$= 2u^{-\frac12} -\frac{2}{5}u^{-\frac52}+C$$
$$= \frac{2}{\sqrt{u}}-\frac{2}{5\sqrt{u^{5}}}+C$$
$$= \frac25\left(\frac{5\sqrt{u^5}-\sqrt{u}}{\sqrt{u}\sqrt{u^{5}}}\right) +C$$
$$= \frac25\left(\frac{5u^2-1}{\sqrt{u^{5}}}\right) +C$$
$$= \frac25\left(\frac{5\sin^2(x)-1}{\sqrt{\sin^{5}(x)}}\right) +C$$
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