Find the antiderivative of (cos 2x)e^sin2x?

Answer 1

Let u= sin2x , then du = 2cos2x dx

So the quation is now (e^u)/2 du

You can factor out the 1/2 to leave the integral of e^u du which is e^u + c

then redistribute the 1/2 to leave e^u + c/2 , but c/2 is any constant, so it becomes just .5e^u + c.

Substitute sin 2x back for u.

The answer is: .5e^(sin2x) + c

Answer 2

use substitution
u=sin2x
du=2cos2x dx

antiderivative of (cos 2x)e^sin2x
=integral of (cos 2x)e^sin2x dx
= integral of1/2 e^u du
=1/2 e^u+C
=1/2 e^sin2x + C

Answer 3

Let u = sin2x Therefore du/dx = 2 cos2x ie du = 2 cos2x dx

Integral(cos 2x)e^sin2x dx
= Integral e^sin2x . cos2x dx
= ½Integral e^sin2x . 2cos2x dx
= ½Integral e^u du
= ½ e^u + c
= ½ e^sin2x + c

Answer 4

let u = sin(2x) then du/dx = 2cos(2x)

then we have cos(2x) (e^u) du/[2cos(2x)

= (1/2)e^u du -------integrate to 1/2 e^(sin(2x)) + C

please check for errors