How do you integrate the indefinite integral of∫(sin2x/sinx)dx?

Answer 1

sin 2x = 2 sin x cos x
So this is just 2 ∫ cos x dx = 2 sin x + C.

Answer 2

Well the integrand is simply 2sinxcosx/sinx = 2cosx

So it's just the Integral(2cos x dx)

Answer 3

i'm going to us INT quite of the mandatory image. INT(((sin9x)^12)(cos9x)^3)dx = INT(((sin9x)^12)(cos9x)^2)(cos9x)dx= INT((sin9x)^12)(a million-(sin9x)^2)(cos9x)dx permit u=sin9x then du = 9(cos9x)dx so du/9=(cos9x)dx after substitution we get a million/9INT(u^12(a million-u^2))du=a million/9INT(u^12-u^14... a million/9((sin9x)^13/13-(sin9x)^15/15) + C