Question

integration of 1+cos2x/1-cos2x

Answer 1

just change the sign. rest is mostly same

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\large\boxed{\sf{\int \dfrac{1-cos2x}{1+cos2x}dx}}$

♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{\int \dfrac{1-\cos \left(2x\right)}{1+\cos \left(2x\right)}dx=\tan \left(x\right)-x+C}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\dfrac{1-\cos (2 x)}{1+\cos (2 x)}=\dfrac{2}{1+\cos (2 x)}-1$

$=\int \dfrac{2}{1+\cos \left(2x\right)}-1dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\int \dfrac{2}{1+\cos \left(2x\right)}dx-\int \:1dx$

$=\tan (x)-x$

$\sf{Add\:a\:constant\:to\:the\:solution}$

$\large\boxed{\sf{=\tan \left(x\right)-x+C}}$