∫ sin2x/√(1+cos^2x) dx
u = 1+cos^2x
du = -2cosx sinx
...
∫ x/[(x^2-1) ln(x^2-1)] dx
u = ln(x^2-1)
du = 1/(x^2-1) * 2x dx
...
Integration:
integral (sin2x / rad (1+cos^2 x) ) dx=?
integral (x / (x^2 -1)(ln (x^2 -1)) ) dx=?
1 answer