Calculus Question!-doubt(a simple one)

=2cos2x/2cos²2x

=(1/cos2x)dx

take tanx=t => sec²xdx=dt

=>(1+t²)dx=dt

then cos2x=1-t²/(1+t²)

so it reduces to

dt/(1-t²)

=1/2(dt/1+t +dt/(1-t))

y 4 all that steps :
u can do it easily::

\int 2(cos2x)/2(cos^{2}2x)\delta x =\int 1/cos2x \delta x =\int sec2x dx =1/2(log\left|sec2x+tan2x \right|)

think its correct

Sorry in denominator it is not 2-2sin²x , but 2 - sin²x

Some one answer please!!It luks bit difficult

u can wrte d numerator as 8-4sin^2x -6

so ur int reduces to

4 - 6/(2-sin^2x)
nw d second one is easy write tan(x/2)=t
or
multiply n divide by sec^2x
then ur second part bcums

6sec^2x/(2sec^2x-tan^2x)
write tanx=t

u r done ;)

sir it is sin²2x

but in numerator u will get (-2 * sin 2x * cos 2x * 2), can u post the soln. pls

arey i said put sin2x as t n not sin²2x