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

Question

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

Elijah Abram · Accepted Answer

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

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#

Cameron Alexander · Answer

See below.

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#

Evelyn Agard · Answer

undefined To integrate $ \int e^x \cos(x^2) \, dx $, we can use the method of integration by parts. Let $ u = e^x $ and $ dv = \cos(x^2) \, dx $. Then, $ du = e^x \, dx $ and $ v = \int \cos(x^2) \, dx $.

The integral $ \int \cos(x^2) \, dx $ does not have a elementary antiderivative. Therefore, we need to leave it as is.

Applying the integration by parts formula $ \int u \, dv = uv - \int v \, du $, we get:

$$ \int e^x \cos(x^2) \, dx = e^x \int \cos(x^2) \, dx - \int \left( \int \cos(x^2) \, dx \right) e^x \, dx $$

So, the integral becomes:

$$ \int e^x \cos(x^2) \, dx = e^x \int \cos(x^2) \, dx - \int e^x \left( \int \cos(x^2) \, dx \right) \, dx $$

This is as simplified as we can get. The integral cannot be expressed in terms of elementary functions.