Helen Lewis

2022-01-05

How to calculate following integration?
$\int \left(\sqrt{\mathrm{tan}x}+\sqrt{\mathrm{cot}x}\right)dx$

kaluitagf

$I=\int \left(\sqrt{\mathrm{tan}x}+\sqrt{\mathrm{cot}x}\right)dx$
$=\int \frac{\mathrm{sin}x+\mathrm{cos}x}{\sqrt{\mathrm{sin}x\mathrm{cos}x}}dx$
Putting
$I=\int \frac{\sqrt{2}du}{\sqrt{1-{u}^{2}}}=\sqrt{2}\mathrm{arcsin}u+C=\sqrt{2}\mathrm{arcsin}\left(\mathrm{sin}x-\mathrm{cos}x\right)+C$
where C is an arbitrary constant for indefinite integral.

alexandrebaud43

Substiute $\mathrm{tan}\left(x\right)=u={e}^{2t}$:
$\int \left(\sqrt{\mathrm{tan}\left(x\right)}+\sqrt{\mathrm{cot}\left(x\right)}\right)dx=\int \left({u}^{\frac{1}{2}}+{u}^{-\frac{1}{2}}\right)\frac{du}{1+{u}^{2}}$
$=\int \frac{{u}^{\frac{1}{2}}+{u}^{-\frac{1}{2}}}{u+{u}^{-1}}\frac{du}{u}$
$=\int \frac{\text{cosh}\left(t\right)}{\text{cosh}\left(2t\right)}2dt$
$=\int \frac{2}{\text{cosh}\left(2t\right)}d\text{sinh}\left(t\right)$
$=\int \frac{\sqrt{2}}{1+2{\text{sinh}}^{2}\left(t\right)}d\sqrt{2}\text{sinh}\left(t\right)$
$=\sqrt{2}\mathrm{arctan}\left(\sqrt{2}\text{sinh}\left(t\right)\right)+C$
$=\sqrt{2}\mathrm{arctan}\left(\frac{\sqrt{\text{tanh}\left(x\right)}-\sqrt{\mathrm{cot}\left(x\right)}}{\sqrt{2}}\right)+C$

karton

Substitute $\mathrm{tan}x={t}^{2}$
$\int \sqrt{\mathrm{tan}x}+\sqrt{\mathrm{cot}x}dx=2\int \frac{{t}^{2}+1}{{t}^{4}+1}dt=2\int \frac{1+\frac{1}{{t}^{2}}}{\left(t-\frac{1}{t}{\right)}^{2}+2}dt$
Now make a $t-\frac{1}{t}$ substitution and we get the answer.

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