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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.

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    $\begingroup$ is $$\sin^{1/2}(x)=\sqrt{\sin(x)}$$? $\endgroup$ Mar 21, 2017 at 15:54
  • 3
    $\begingroup$ Try $u = \tan(x)$. $\endgroup$ Mar 21, 2017 at 15:56

2 Answers 2

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

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

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

Set $\tan x=u$

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