# Find solution to exercise:What is integral of (cos x)*(e^2x)?

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let f(x) = cosx*e^2x

We need to find the integral of f(x).

==> intg f(x) = intg cosx * e^2x dx

We will use partial integration to solve.

Let us assume that:

u= e^2x ==> du = 2e^2x dx

dv = cosx dx ==> v = sinx

==> intg udv = u*v - intg vdu

= sinx*e^2x + 2 intg sinx*e^2x dx.............(1)

Now we will apply the rule again.

Let u = e^2x ==> du 2e^2x

dv = sinx dx ==> v = -cosx

==> intg sinx*e^2x dx = -cosx*e^2x +2 intg cosx*e^2x dx

==> But we know that intg cosx*e^2x dx = intg udv

==> intg udv = sinx*e^2x + 2[ cosx*e^2x -2intg udv]

==> intg udv = sinx*e^2x + 2cosx*e^2x -4intg udv.

==> 5intg udv = sinx*e^2x +2cosx*e^2x

**==> intg udv = e^2x ( sinx+2cosx) / 5 **

To find integral cosx*e^2x dx.

Let I = cosx*e^2x dx.

I = cosx*Int e^2x dx - Int {(cosx)' Int e^xdx } dx.

I = cosx (e^2x)/2 - (1/2) Int {-sinx* e^2x} dx

I = (1/2) cosxe^2x + (1/2) Int {sinx e^2x dz}

I = (1/2) cosx e^2x +(1/2){sinx (e^2x)/2 - (1/2) I}

I = (1/2) cosx e^2x + (1/4) sinx 2^2x - I/4

(1+1/4)I = (1/4){2cosx + sinx}e^(2x).

(5/4)I = (1/4){2cosx + sinx}e^(2x).

I = (4/5)(1/4) {2cosx + sinx}e^(2x).

I = (1/5){2cosx + sinx}e^(2x).

Therefore Int cosx e^2x dx = (1/5){2cosx + sinx}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**