Integral of exp(x)*cos(x) by parts?

Question

I have been asked to find the integral of exp(x)*cos(x) by parts, can anyone help me out? Unless I am doing it wrong, exp(x)*cos(x) alternates to exp(x)*cos(x) minus the integral of exp(x)*sin(x) and back according to integration by parts?

Answer 1

The following method is based on

integration by parts but is much shorter.

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Note : ∫ ℯˣ. [ ƒ(x) + ƒ'(x) ] dx = ℯˣ. ƒ(x) ................ (1)

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........... ∫ ℯˣ. cos x dx



= (1/2) ∫ ℯˣ. ( 2 cos x ) dx



= (1/2) ∫ ℯˣ. [ ( cos x + sin x ) + ( -sin x + cos x ) ] dx ... Note This Step !



= (1/2) ∫ ℯˣ. [ ƒ(x) + ƒ'(x) ] dx, ........... ƒ(x) = cos x + sin x



= (1/2) ℯˣ. ƒ(x) + C ......................... from (1)



= (1/2). ℯˣ. ( cos x + sin x ) + C .............................................. Ans.

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Happy To Help !

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Answer 2

You preform integration by parts twice in this case

choosing :



u = cos(x) du = -sin(x)

and

dv = exp(x) v = exp(x) dx



Then putting these values together in the integration by parts equation you get:

integral[exp(x)*cos(x)dx = exp(x)*cos(x) + integral[exp(x)*sin(x)



you see then you have another integral that you need to use by parts again with. Do this the same way as before using the exponent as dv and v. Eventually you should get the same integral on both sides and can add them. With a little shuffling around of numbers you should get the original integral on one side and an equation with no integrals on the other plus C (arbitrary constant of integration). Hope this helps. Good luck!

Answer 3

Integration by parts is much easier to do in terms of functions and derivatives rather than differentials, yet the first answerer has chosen the more complicated way.



Integrate the original integrand by parts:

∫ ℮˟cosx dx

Let f'(x) = cosx

f(x) = sinx

Let g(x) = ℮˟

g'(x) = ℮˟

∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dx

∫ ℮˟cosx dx = ℮˟sinx - ∫ ℮˟sinx dx



Integrate the new integrand by parts:

∫ ℮˟sinx dx

Let f'(x) = sinx

f(x) = -cosx

Let g(x) = ℮˟

g'(x) = ℮˟

∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dx

∫ ℮˟sinx dx = -℮˟cosx + ∫ ℮˟cosx dx



Put it all together to solve for the original integral:

∫ ℮˟cosx dx = ℮˟sinx - ∫ ℮˟sinx dx

∫ ℮˟cosx dx = ℮˟sinx - (-℮˟cosx + ∫ ℮˟cosx dx)

∫ ℮˟cosx dx = ℮˟sinx + ℮˟cosx - ∫ ℮˟cosx dx

2 ∫ ℮˟cosx dx = ℮˟sinx + ℮˟cosx

2 ∫ ℮˟cosx dx = ℮˟(sinx + cosx)

∫ ℮˟cosx dx = ℮˟(sinx + cosx) / 2 + C