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OpenStudy (anonymous):

Using the substitution t=tan(x/2), find dx/cosx dx between(0,pi/3)

OpenStudy (anonymous):

$\int\limits_{0}^{\Pi/3}dx/cosx$

OpenStudy (jwt625):

$\int\limits_{?}^{?}(dx/\cos(x))=\ln(\sec(x)+\tan(x))$

OpenStudy (jwt625):

the answer is $\ln (2+\sqrt{3})$

OpenStudy (anonymous):

thanks i'll try work it out now

OpenStudy (jwt625):

you can check it out

OpenStudy (zarkon):

if you have to use that substitution then you will get the following... $$\int\frac{1}{\cos(x)}dx$$ $$=\int\frac{1+t^2}{1-t^2}\frac{1}{1+t^2}dt=2\int\frac{1}{1-t^2}dt$$ $$=2\int\frac{1}{(1-t)(1+t)}dt=2\int\left(\frac{1}{2(1+t)}+\frac{1}{2(1-t)}\right)$$ $$=\ln(1+t)-\ln(1-t)+c=\ln\left(\frac{1+t}{1-t}\right)+c$$ $$=\ln\left(\frac{1+t}{1-t}\frac{1+t}{1+t}\right)+c=\ln\left(\frac{(1+t)(1+t)}{1-t^2}\right)+c$$ $$=\ln\left(\frac{2t+1+t^2}{1-t^2}\right)+c=\ln\left(\frac{2t}{1-t^2}+\frac{1+t^2}{1-t^2}\right)+c$$ note that $$\tan(x)=\tan(x/2+x/2)= \frac{2\tan x/2}{1 -\tan^2(x/2)}=\frac{2t}{1-t^2}$$ hence we have $$\ln\left(\tan(x)+\sec(x)\right)+c$$

OpenStudy (zarkon):

$$\int\frac{1}{\cos(x)}dx$$ $$=2\int\frac{1+t^2}{1-t^2}\frac{1}{1+t^2}dt=2\int\frac{1}{1-t^2}dt$$ $$=2\int\frac{1}{(1-t)(1+t)}dt=2\int\left(\frac{1}{2(1+t)}+\frac{1}{2(1-t)}\right)$$ $$=\ln(1+t)-\ln(1-t)+c=\ln\left(\frac{1+t}{1-t}\right)+c$$ $$=\ln\left(\frac{1+t}{1-t}\frac{1+t}{1+t}\right)+c=\ln\left(\frac{(1+t)(1+t)}{1-t^2}\right)+c$$ $$=\ln\left(\frac{2t+1+t^2}{1-t^2}\right)+c=\ln\left(\frac{2t}{1-t^2}+\frac{1+t^2}{1-t^2}\right)+c$$ note that $$\tan(x)=\tan(x/2+x/2)= \frac{2\tan x/2}{1 -\tan^2(x/2)}=\frac{2t}{1-t^2}$$ hence we have $$\ln\left(\tan(x)+\sec(x)\right)+c$$ I left off a 2 in the second line

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