e^ (2x) sinx dx?

Answer 1

Integral ( e^(2x) sin(x) dx )

You have to use parts to solve this.

Let u = e^(2x). dv = sin(x) dx
du = 2e^(2x) dx. v = -cos(x) dx

= -e^(2x)cos(x) - Integral ( -2e^(2x)cos(x) dx )

Factoring out a -2 gives us

= -e^(2x)cos(x) + 2*Integral (e^(2x)cos(x) dx)

Now, use parts again.

Let u = e^(2x). dv = cos(x) dx
du = 2e^(2x) dx. v = sin(x)

= -e^(2x)cos(x) + 2*[e^(2x)sin(x) - Integral (2e^(2x)sin(x) dx)]

Distribute the 2 over the brackets.

= -e^(2x)cos(x) + 2e^(2x)sin(x) - 2*Integral (2e^(2x)sin(x) dx)

Factor the 2 out of the integral.

= -e^(2x)cos(x) + 2e^(2x)sin(x) - 4*Integral(e^(2x)sin(x) dx)

You may have notice that we went in a circle; that's not the case, however, because to the left of the equal sign, we have
Integral (e^(2x)sin(x) dx), i.e.

Integral (e^(2x)sin(x) dx) = -e^(2x)cos(x) + 2e^(2x)sin(x) - 4*Integral(e^(2x)sin(x) dx)

At this point, what we're going to do is *treat* the integral like a variable; note that we have one of them on the left hand side, and -4 of them on the right hand side. Therefore, we *add* 4 times the Integral on both sides.

Integral (e^(2x)sin(x) dx) + 4*Integral (e^(2x)sin(x) dx) = -e^(2x)cos(x) + 2e^(2x)sin(x)

We can now essentially "group" the two integrals the same way we add y + 4y; we get 5y, so we have

5*Integral (e^(2x)sin(x) dx) = -e^(2x)cos(x) + 2e^(2x)sin(x)

Multiply both sides by (1/5), to obtain

Integral (e^(2x)sin(x) dx) = (-1/5)e^(2x)cos(x) + (2/5)e^(2x)sin(x)

And don't forget to add the constant C.

Integral (e^(2x)sin(x) dx) = (-1/5)e^(2x)cos(x) + (2/5)e^(2x)sin(x) + C

Answer 2

nice to hear, now what would you like to know?