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I don't know why $$\int_0^{\pi/2}\tan\frac{x}2\ dx= \ln 2.$$ How can i solve this to get that answer?

Daniel W. Farlow
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Soroush
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    What is the antiderivative of $\tan{x}$? Start with that. – Michael Dyrud Aug 25 '15 at 14:01
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    Please edit the question to include additional background and motivation. Also, please include a description of what you have already tried. These things help make the question more compelling, help others find it, and help others write more useful answers. – Carl Mummert Aug 27 '15 at 11:07

4 Answers4

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If we set $x=2\arctan t$ we have: $$ \int_{0}^{\pi/2}\tan\left(\frac{x}{2}\right)\,dx = \int_{0}^{1}\frac{2t}{1+t^2}\,dt = \left.\log(1+t^2)\right|_{0}^{1}=\color{red}{\log 2}$$ as wanted.

Jack D'Aurizio
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  • +1 I was about to post an answer with the Weierstrass sub (which allows you to "see it" immediately), but you beat me to the punch. :) – Deepak Aug 25 '15 at 14:23
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Hint

$tan(x/2) = \frac {sin(x/2)}{cos(x/2)}=-2\frac {(cos(x/2))'}{cos(x/2)}-=-2(ln(cos(x/2)))'$

Therefore:

$\int_{0}^{\pi/2}tan(x/2) dx=-2ln(cos(\pi/4)) + 2ln(cos(0))=-2ln(1/\sqrt 2 )=ln2$

2

$$\int_0^{\frac{\pi}{2}}\tan\left(\frac{x}{2}\right) dx$$ Using $u$-substitution, we have $$u=\frac{x}{2}\Rightarrow 2\ du=dx$$ So now $$2\int_0^{\frac{\pi}{4}}\tan u\ du$$ $$=-2\ln\left|\cos \frac{\pi}{4}\right|+2\ln\left|\cos 0\right|$$ $$=-2\ln\frac{1}{\sqrt2}+2\ln 1$$ $$=-2\ln 1+2\ln \sqrt2+2\ln 1$$ $$=2\ln \sqrt2$$ $$=\ln 2$$

k170
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Notice $$\int_{0}^{\pi/2}\tan\frac{x}{2}dx=\left[2\ln\left|\sec\frac{x}{2}\right|\right]_{0}^{\pi/2}$$ $$=2[\ln|\sec\frac{\pi}{4}|-\ln|\sec 0|]$$ $$=2[\ln|\sqrt 2|-\ln|1|]$$ $$=2\ln|\sqrt 2|=2\frac{1}{2}\ln 2=\ln 2$$

Harish Chandra Rajpoot
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