Question

How do you integrate `int e^x cos x^2 dx`?

Answer 1

`int \ e^x (cosx)^2 \ dx = 1/2 e^x + 1/5 e^xsin2x +1/10 e^xcos2x + C`

Explanation:

Assuming that we seek:

`I = int \ e^x (cosx)^2 \ dx`

Then, using the identity `cos2theta -= 2cos^2theta-1` we can rewrite the integral as:

`I = int \ e^x (1+cos2x)/2 \ dx`
`\ \ = 1/2 \ int \ e^x (1+cos2x) \ dx`
`\ \ = 1/2 \ {int \ e^x \ dx + int \ e^x cos2x \ dx }` ..... [A]

The first integral is trivial, for the second we apply integration by parts:

Let `{ (u,=e^x, => (du)/dx,=e^x), ((dv)/dx,=cos2x, => v,=1/2 sin2x ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We have:

`int \ e^xcos2x \ dx = e^x1/2sin2x-int \ 1/2sin2x \ e^x \ dx`
`" " = 1/2 e^xsin2x - 1/2 \ int \ e^xsin2x \ dx` ..... [B]

Now if we perform a second application of Integration By Parts:

Let `{ (u,=e^x, => (du)/dx,=e^x), ((dv)/dx,=sin2x, => v,=-1/2 cos2x ) :}`

Then plugging into the IBP formula, we have:

`int \ e^xsin2x \ dx = e^x(-1/2cos2x)-int \ (-1/2cos2x) \ e^x \ dx`
`" " = -1/2 e^xcos2x + 1/2 \ int \ e^xcos2x \ dx` ..... [C]

Substituting result [C] into [B] we have:

`int \ e^xcos2x \ dx = 1/2 e^xsin2x - 1/2 {-1/2 e^xcos2x + 1/2 \ int \ e^xcos2x \ dx}`

`:. int \ e^xcos2x \ dx = 1/2 e^xsin2x +1/4 e^xcos2x -1/4 \ int \ e^xcos2x \ dx`

`:. 5/4 \ int \ e^xcos2x \ dx = 1/2 e^xsin2x +1/4 e^xcos2x`

`:. int \ e^xcos2x \ dx = 2/5 e^xsin2x +1/5 e^xcos2x`

Now, substituting this result into [A] and integrating we get:

`I = 1/2 \ {e^x + 2/5 e^xsin2x +1/5 e^xcos2x} + C`
`\ \ = 1/2 e^x + 1/5 e^xsin2x +1/10 e^xcos2x + C`

Answer 2

See below.

Explanation:

Assuming the question reads

`int e^x cos^2x dx = 1/2int e^x(1+cos(2x))dx`

Now `int e^x cos(2x) dx = "Re"[int e^x e^(2ix)dx] = "Re"[int e^((2i+1)x) dx]`

Now

`int e^((2i+1)x)dx = 1/(2i+1) e^((2i+1)x)+C = e^x/(2i+1)(cos (2x)+i sin(2x))+C`

and `"Re"[ e^x/(2i+1)(cos (2x)+i sin(2x))+C] =1/5 e^x (Cos(2 x) + 2 Sin(2 x))+C_1`

so finally

`int e^x cos^2x dx =e^x/2(1+1/5 (Cos(2 x) + 2 Sin(2 x)))+C_2`