Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

thanks i'll try work it out now

Answer 5

you can check it out

Answer 6

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


$$\ln\left(\tan(x)+\sec(x)\right)+c$$

Answer 7

$$\int\frac{1}{\cos(x)}dx$$

$$=2\int\frac{1+t^2}{1-t^2}\frac{1}{1+t^2}dt=2\int\frac{1}{1-t^2}dt$$

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

$$=\ln(1+t)-\ln(1-t)+c=\ln\left(\frac{1+t}{1-t}\right)+c$$

$$=\ln\left(\frac{1+t}{1-t}\frac{1+t}{1+t}\right)+c=\ln\left(\frac{(1+t)(1+t)}{1-t^2}\right)+c$$

$$=\ln\left(\frac{2t+1+t^2}{1-t^2}\right)+c=\ln\left(\frac{2t}{1-t^2}+\frac{1+t^2}{1-t^2}\right)+c$$

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