How to solve this integral? I solved it and got -ln(2)+ln(5/4).

Question

#int_0^(pi/3)sin(2x)/(1+cos^2x)dx#

909 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-26T19:45:23

#\ln2-\ln(5/4)#

#\int_0^{\pi/3} \frac{\sin(2x)}{1+\cos^2x}\ dx#

#=\int_0^{\pi/3} \frac{2\sinx\cos x}{1+1-\sin^2x}\ dx#

#=\int_0^{\pi/3} \frac{2\sinx\cos x}{2-\sin^2x}\ dx#

Let #\sin x=t\implies \cos x\ dx=dt#

#=\int_0^{\sqrt3/2}\frac{2t\dt }{2-t^2}#

#=-\int_0^{\sqrt3/2}\frac{-2t\dt }{2-t^2}#

#=-\int_0^{\sqrt3/2}\frac{d(2-t^2) }{2-t^2}#

#=-(\ln|2-t^2|)_0^{\sqrt3/2}#

#=-\ln|2-3/4|+\ln|2|#

#=-\ln(5/4)+\ln 2#

#=\ln 2-\ln(5/4)#