Solve $\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx$
With $u=\tan(x)$, you get $du=\frac{dx}{\cos^2(x)}$ : $$\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx=\int \frac{u^2+1}{\sqrt u}du=\int u^{\frac{3}{2}}du+\int \frac{1}{\sqrt u}du\\=\frac{2u^{\frac{5}{2}}}{5}+2\sqrt u + C=\frac{2\tan^{\frac{5}{2}}(x)}{5}+2\sqrt{\tan(x)} + C$$
Generalization:
$$\int\dfrac{dx}{\sin^ax\cos^{2b-a}x}=\int\dfrac{(1+\tan^2x)^{b-1}}{\tan^ax}\sec^2x\ dx$$
Set $\tan x=u$