Solve $\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx$

So it's given this indefinite integral $$\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx$$

Is there anyone could solve this integral? Thanks in advance.

• is $$\sin^{1/2}(x)=\sqrt{\sin(x)}$$? Mar 21, 2017 at 15:54
• Try $u = \tan(x)$. Mar 21, 2017 at 15:56

With $u=\tan(x)$, you get $du=\frac{dx}{\cos^2(x)}$ : $$\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx=\int \frac{u^2+1}{\sqrt u}du=\int u^{\frac{3}{2}}du+\int \frac{1}{\sqrt u}du\\=\frac{2u^{\frac{5}{2}}}{5}+2\sqrt u + C=\frac{2\tan^{\frac{5}{2}}(x)}{5}+2\sqrt{\tan(x)} + C$$

• Sorry I made a typo. It's 7/2. Mar 21, 2017 at 16:00
• The substitution still works for the changed problem, it's just a little more complicated. Mar 21, 2017 at 16:02
• here we go :) ${}$ Mar 21, 2017 at 16:06
• Hope you don't mind the line-breaking. Mar 21, 2017 at 20:12
• no problem ;) ${}$ Mar 21, 2017 at 20:29

Generalization:

$$\int\dfrac{dx}{\sin^ax\cos^{2b-a}x}=\int\dfrac{(1+\tan^2x)^{b-1}}{\tan^ax}\sec^2x\ dx$$

Set $\tan x=u$