MHB Problem of the Week #122 - July 28th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Evaluate $\displaystyle\int\sqrt{\tan x}\,dx$.

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Hints: [sp]Start with the substitution $t^2=\tan x$.

Also note that $t^4+1 = (t^2+1)^2-(\sqrt{2}t)^2 = (t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)$.[/sp]

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Kiwi. You can find their solution below.

[sp]let

t^2=\tan (x), then

2tdt=\frac {dx}{\cos^2(x)}=\frac{(\cos^2(x)+\sin^2(x))dx}{\cos^2(x)}=(1+\tan^2(x))dx =(t^4+1)dx=((t^2+1)^2-(\sqrt2t)^2)dx

\therefore 2tdt=(t^2+\sqrt 2 t +1)(t^2-\sqrt 2 t +1)dx

\therefore tdx=\frac {2t^2dt}{(t^2+\sqrt 2 t +1)(t^2-\sqrt 2 t +1)}=\frac {tdt}{\sqrt 2 (t^2-\sqrt 2 t +1)}-\frac {tdt}{\sqrt 2 (t^2+\sqrt 2 t +1)}

So
\int \sqrt{\tan(x)}dx
=\frac{1}{\sqrt 2}\int \frac {(t-\sqrt 2)dt}{(t^2-\sqrt 2 t +1)}-\frac{1}{\sqrt 2}\int \frac {(t+\sqrt 2)dt}{(t^2+\sqrt 2 t +1)}+\frac{1}{\sqrt 2}\int \frac {\sqrt 2 dt}{(t^2-\sqrt 2 t +1)}+\frac{1}{\sqrt 2}\int \frac {(\sqrt 2)dt}{(t^2+\sqrt 2 t +1)}

=\frac 1{\sqrt 2} \ln \left(\frac{t^2-\sqrt 2 t +1}{t^2+\sqrt 2 t +1}\right)+\int \frac {dt}{(t^2-\sqrt 2 t +1)}+\int \frac {dt}{(t^2+\sqrt 2 t +1)}

=\frac 1{\sqrt 2} \ln \left(\frac{t^2-\sqrt 2 t +1}{t^2+\sqrt 2 t +1}\right)+\int \frac {dt}{(t-\frac{1}{\sqrt 2})^2+ (\frac{1}{\sqrt 2})^2)}+\int \frac {dt}{(t+\frac{1}{\sqrt 2})^2+ (\frac{1}{\sqrt 2})^2)}

Now let \frac 1 {\sqrt 2}\tan(s_1)=t-\frac 1{\sqrt 2} and \frac 1 {\sqrt 2}\tan(s_2)=t+\frac 1{\sqrt 2} then

\frac 1{\sqrt 2} \frac{ds_*}{\cos^2(s_*)}=dt therefore:

\int \sqrt{\tan(x)}dx

=\frac 1{\sqrt 2} \ln \left(\frac{t^2-\sqrt 2 t +1}{t^2+\sqrt 2 t +1}\right)+\frac 1{2\sqrt 2}\int \frac {ds_1}{(\tan^2(s_1)+ 1)\cos^2(s_1)}+\frac 1{2\sqrt 2}\int \frac {ds_2}{(\tan^2(s_2)+ 1)\cos^2(s_2)}

=\frac 1{\sqrt 2} \ln \left(\frac{t^2-\sqrt 2 t +1}{t^2+\sqrt 2 t +1}\right)+\frac {s_1+s_2}{2\sqrt 2}+C

=\frac 1{\sqrt 2} \ln \left(\frac{t^2-\sqrt 2 t +1}{t^2+\sqrt 2 t +1}\right)+\frac {\tan^{-1}(\sqrt 2 t-1)+\tan^{-1}(\sqrt 2 t+1)}{2\sqrt 2}+C

=\frac 1{\sqrt 2} \ln \left(\frac{\tan(x)-\sqrt {2 \tan(x)} +1}{\tan(x)+\sqrt {2 \tan(x)} +1}\right)+\frac {\tan^{-1}(\sqrt{2\tan(x)}-1)+\tan^{-1}( \sqrt{2\tan(x)}+1)}{2\sqrt 2}+C[/sp]