Please help with integrals?

Question

evaluate the integral

∫ e^ -θ cos2θdθ

Answer 1

Integration by parts: ∫ u dv = uv - ∫ v du

let u = e^ (-θ)

then du = -(e^(-θ)) dθ

and let dv = cos(2θ) dθ

then v =  sin(2θ)/2

hence

1] ∫ e^ (-θ) cos (2θ) dθ = e^ (-θ)*sin(2θ)/2 + 1/2 ∫ e^ (-θ) sin(2θ)dθ

Now apply ∫ by parts to

∫ e^ (-θ) sin(2θ) dθ

let u = e^ (-θ)

then du = -(e^(-θ)) dθ

and let dv = sin(2θ) dθ

then v =  - cos(2θ)/2

hence

2] ∫ e^ (-θ) sin(2θ) dθ = - e^ (-θ)*cos(2θ)/2 - 1/2 ∫ e^ (-θ) cos(2θ)dθ

Substitute 2] into 1]

∫ e^ (-θ) cos (2θ) dθ = e^ (-θ)*sin(2θ)/2 + 1/2 * [- e^ (-θ)*cos(2θ)/2 - 1/2 ∫ e^ (-θ) cos(2θ)dθ]

do the algebra and you obtain...

5/4 * ∫ e^ (-θ) cos (2θ) dθ = e^ (-θ)*sin(2θ)/2 - e^ (-θ)*cos(2θ)/4

so...

∫ e^ (-θ) cos (2θ) dθ = 1/5 * (2 e^ (-θ)*sin(2θ) - e^ (-θ)*cos(2θ))

Simplified answer:

∫ e^ (-θ) cos (2θ) dθ = 1/5 * e^ (-θ) * (2*sin(2θ) - cos(2θ)) + c