We need to find the integral of (cos x)*(e^2x)

Here integration by parts is the method to be used.

Int [ u dv] = v*u - Int[ v du]

u = cos x, du = -sin x

dv = e^2x , v = e^2x / 2

Int[(cos x)*(e^2x)] = cos x*e^2x / 2 + Int[ sin x*e^2x]/2

Again use integration by parts for Int[ sin x*e^2x]/2

u = sin x , du = cos x

dv = e^2x, v = e^2x / 2

Int[ sin x*e^2x]/2 = sin x*e^2x / 4 - Int [cos x e^2x] / 4

So we get Int[(cos x)*(e^2x)] = (4/5)[cos x*e^2x / 2 +sin x*e^2x / 4]

=> (2/5)*cos x*e^2x + (1/5)*sin x*e^2x + C

**The integral of (cos x)*(e^2x) is (2/5)*cos x*e^2x + (1/5)*sin x*e^2x + C**

We'll apply integration by parts:

Int udv = uv - Int vdu

We'll put u = e^2x

We'll differentiate:

du = 2e^2x

We'll put dv = cos x*dx

v = Int cos x dx = sin x

Int udv = (e^2x)*(sin x) - 2Int (e^2x)*(sin x)

We'll apply integration by parts again:

We'll put u = e^2x

We'll differentiate:

du = 2e^2x

We'll put dv = sin x*dx

v = - cos x

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

We'll remove the brackets:

Int udv = (e^2x)*(sin x) + 2(e^2x)(cos x) - 4 Int udv

We'll add 4Int udv both sides:

5Int udv = (e^2x)*(sin x) + 2(e^2x)(cos x)

**Int udv = (e^2x)*(sin x + 2 cos x)/5 + C**