How do you find the integral of #(cosx)(e^x)#?

Question

Isaiah Allen · Accepted Answer

# int \ cosx \ e^x \ dx = 1/2e^x(cosx + sinx) + C #

Let:

# I = int \ cosx \ e^x \ dx #

We may apply integration by components:

Let # { (u,=cosx, => (du)/dx=-sinx), ((dv)/dx,=e^x, => v=e^x ) :}#

After that, entering the IBP formula:

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

gives us

# int \ (cosx)(e^x) \ dx = (cosx)(e^x) - int \ (e^x)(-sinx) \ dx # # :. I = e^xcosx + int \ e^x \ sinx \ dx # .... [A]

At first it appears as if we have made no progress, as now the second integral is similar to #I#, having exchanged #cosx# for #sinx#, but if we apply IBP a second time then the progress will become clear:

Let # { (u,=sinx, => (du)/dx=cosx), ((dv)/dx,=e^x, => v=e^x ) :}#

Subsequently, entering the IBP formula provides us with:

# int \ (sinx)(e^x) \ dx = (sinx)(e^x) - int \ (e^x)(cosx) \ dx # # :. int \ e^x \ sinx \ dx = e^xsinx - I #

When we enter this outcome into [A], we obtain:

# I = e^xcosx + e^xsinx - I + A #

# :. 2I = e^xcosx + e^xsinx + A # # :. I = 1/2(e^xcosx + e^xsinx) + C #

Christopher Agresta · Answer

#inte^x cosx dx = (e^x(cosx+sinx))/2 +C#

Integrate by parts:

#int e^x cosx dx = int cosx d(e^x) = e^xcosx - int e^x d(cosx) = e^x cosx +int e^xsinxdx#

Now integrate again by parts:

#int e^x sinx dx = int sinx d(e^x) = e^xsinx - int e^x d(sinx) = e^x sinx -int e^xcosxdx#

Substituting this equality in the first one we have:

#int e^x cosx dx = e^xcosx+e^xsinx - int e^xcosxdx#

The integral now appears on both sides of the equation and we can now solve for it:

#2int e^x cosx dx = e^xcosx+e^xsinx +C#

and finally:

#inte^x cosx dx = (e^x(cosx+sinx))/2 +C#

Ezra Adams · Answer

Recursively use integration by parts.
After 2 iterations, the negative of the integral will be on the right.
Subtract form to both sides and divide by 2.

Integration by parts: #intudv = uv - intvdu# let #u = cos(x)#, then, #dv = e^xdx#, #du = -sin(x)dx#, and #v = e^x# Substitute into the formula: #intcos(x)e^xdx=cos(x)e^x-int-sin(x)e^xdx# #intcos(x)e^xdx=cos(x)e^x+intsin(x)e^xdx" [1]"# Integration by parts: #intudv = uv - intvdu# let #u = sin(x)#, #dv = e^xdx#, then, #du = cos(x)dx#, and #v = e^x# Substitute into the formula: #intsin(x)e^xdx = sin(x)e^x - intcos(x)e^xdx# Substitute into equation [1]: #intcos(x)e^xdx=cos(x)e^x+sin(x)e^x - intcos(x)e^xdx# Add #intcos(x)e^xdx# to both sides: #2intcos(x)e^xdx=cos(x)e^x+sin(x)e^x# Divide by 2: #intcos(x)e^xdx=1/2cos(x)e^x+1/2sin(x)e^x# Don't forget the constant: #intcos(x)e^xdx=1/2cos(x)e^x+1/2sin(x)e^x+ C#

Aurora Alvarado · Answer

undefined To find the integral of $ \cos(x) \cdot e^x $, you can use integration by parts. Let $ u = \cos(x) $ and $ dv = e^x \, dx $. Then, $ du = -\sin(x) \, dx $ and $ v = e^x $. Applying the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

$$ \int \cos(x) \cdot e^x \, dx = \cos(x) \cdot e^x - \int (-\sin(x) \cdot e^x) \, dx $$

$$ = \cos(x) \cdot e^x + \int \sin(x) \cdot e^x \, dx $$

Now, you can use integration by parts again on $ \int \sin(x) \cdot e^x \, dx $, or you can recognize that it's the same integral you started with, but with $ \sin(x) $ instead of $ \cos(x) $. Therefore:

$$ \int \cos(x) \cdot e^x \, dx = \cos(x) \cdot e^x + \sin(x) \cdot e^x + C $$

where $ C $ is the constant of integration.