Integral of sin2x/(1+cos2x) on [0,pi/4] using the reverse chain rule or chain rule backwards.

Whenever we see the derivative of the denominator in the numerator, we know it integrates to the natural log of the denominator!
In this video, we compute the integral of sin(2x)/(1+cos(2x)) using the chain rule backwards, also called the reverse chain rule. In other words, we're recognizing the consequences of the chain rule in an integral, rather than relying on an explicit u-substitution.
We see immediately that the numerator can be modified by a constant to give us the derivative of the denominator, and we quickly find the antiderivative and apply the limits of integration, and we're done!

[@nailedthepotato1941]

those who struggle you can just simplify the equation with trigonometry.
cos(2x) = 2cos^2(x) -1
sin(2x) = 2sin(x)cos(x)
so you get:
2sin(x)cos(x) 2sin(x)cos(x) sin(x)
------------------------- = --------------------------- = ------------ = tan(x)
1+ 2cos^2(x) -1 2*cos(x)*cos(x) cos(x)
the integral of tan(x) is: -ln(abs(cos(x))
-ln(cos(pi/4)) + ln(cos(0)) = -ln(sqrt(2)/2) + ln(1) = ln((sqrt(2)/2))^(-1))
=ln(2/sqrt(2)) = ln(sqrt(2)) = (ln(2))/2

[@VarunChaturvedi-u9q]

Sooo its A(d/dx 1+cos2x) +B= sin2x .....am i right