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December 5th, 2016, 02:39 PM   #1
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Question integral(dx/sqrt(tanx))

Does anyone have any useful ideas how to solve this integral:

integral(dx/sqrt(tanx))

$\displaystyle \int \frac{dx}{\sqrt{tan{x}}}$

I really do not know how to start solving it. Please help me guys!
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December 5th, 2016, 03:20 PM   #2
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Weierstrass substitution $t= \tan \frac{x}2$.
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December 5th, 2016, 08:41 PM   #3
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Make the sub $t^2=\tan x$. You'll end up with

$$2\int\frac{1}{1+t^4}\text{ d}t$$

$1+t^4$ factors as $(t^2+\sqrt2t+1)(t^2-\sqrt2t+1)$

Now use partial fractions with complex numbers ... etc.
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December 8th, 2016, 02:12 AM   #4
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Quote:
Originally Posted by srecko View Post
Does anyone have any useful ideas how to solve this integral:

integral(dx/sqrt(tanx))

$\displaystyle \int \frac{dx}{\sqrt{tan{x}}}$

I really do not know how to start solving it. Please help me guys!
$\displaystyle \begin{align*} \int{ \frac{\mathrm{d}x}{\sqrt{\tan{(x)}}} } &= \int{ \frac{\sec^2{(x)}\,\mathrm{d}x}{\sec^2{(x)}\,\sqrt {\tan{(x)}}} } \\ &= \int{ \frac{ \sec^2{(x)} \, \mathrm{d}x }{ \left[ 1 + \tan^2{(x)} \right] \, \sqrt{ \tan{(x)} } } } \\ &= \int{ \frac{\mathrm{d}u}{\left( 1 + u^2 \right) \,\sqrt{u}} } \textrm{ after substituting } u = \tan{(x)} \implies \mathrm{d}u = \sec^2{(x)}\,\mathrm{d}x \\ &= 2\int{ \frac{\mathrm{d}u}{ \left[ 1 + \left( \sqrt{u} \right) ^4 \right] \, 2\,\sqrt{u} } } \\ &= 2\int{ \frac{\mathrm{d}t}{1 + t^4} } \textrm{ after substituting } t = \sqrt{u} \implies \mathrm{d}t = \frac{\mathrm{d}u}{2\,\sqrt{u}} \end{align*}$

which matches what Greg said. Now follow the rest of his advice
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