My Math Forum integral(dx/sqrt(tanx))
 User Name Remember Me? Password

 Real Analysis Real Analysis Math Forum

 December 5th, 2016, 02:39 PM #1 Member   Joined: Oct 2016 From: Slovenia, Europe Posts: 46 Thanks: 5 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!
 December 5th, 2016, 03:20 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,540 Thanks: 2146 Math Focus: Mainly analysis and algebra Weierstrass substitution $t= \tan \frac{x}2$. Thanks from topsquark and srecko
 December 5th, 2016, 08:41 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,435 Thanks: 870 Math Focus: Elementary mathematics and beyond 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. Thanks from topsquark and srecko
December 8th, 2016, 02:12 AM   #4
Member

Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

Quote:
 Originally Posted by srecko 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

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post MthabisiMhlanga Calculus 1 December 7th, 2014 05:30 PM Chemist@ Calculus 1 November 29th, 2014 09:22 AM aimforthehead Calculus 3 February 3rd, 2013 12:15 PM ungeheuer Calculus 19 January 28th, 2013 10:59 AM Riazy Calculus 3 January 15th, 2011 01:41 PM

 Contact - Home - Forums - Top