Question

int ( 2x / ( 1 - cos 2x) )dx ; anybody ?

Answer 1

help needed guys .
\[\int\limits (2x / ( 1+ \cos 2x)) dx\]

Answer 2

...

Answer 3

?

Answer 4

lol let lalaly finish

Answer 5

@nickhouraney ; okay and @lalaly : no pressure take your time .

Answer 6

\[\int\limits{\frac{2x}{1+\cos(2x)}dx}\]sincee\[\frac{1}{2}(1+\cos(2x))=\cos^2x\]so integration becomes\[\frac{1}{2}\times 2 \int\limits{\frac{x}{\frac{1}{2}(1+\cos(2x))}}dx\]
\[=\int\limits{\frac{x}{\cos^2x}}dx\]\[=\int\limits{xsec^2x}dx\]
now solve by parts
u=x dv=sec^2xdx
du=dx v=tanx

so integral becomes\[xtan(x)-\int\limits{\tan(x)dx}\]
do u know how to integrate tan(x)

Answer 7

Forgot

Answer 8

Sorry

Answer 9

ok then\[\tan(x)=\frac{\sin(x)}{\cos(x)}\]
so\[\int\limits{\tan(x)}dx=\int\limits{\frac{\sin(x)}{\cos(x)}}dx\]now let u=cosx
du=-sin(x)dx

substitute in the integral\[\int\limits{\frac{-du}{u}}\]\[=-\ln(u)\]substitute u=cos(x)\[\int\limits{\tan(x)dx}=-\ln(\cos(x))+C\]
so\[\int\limits{\frac{2x}{(1+\cos(2x)}}dx=xtan(x)-\ln(\cos(x))+c\]

Answer 10

Thank you very much ; miss high school .

Answer 11

ur welcome