How do you integrate #e^x*cos(x)#?

Question

Emma Aldridge · Accepted Answer

#e^x/2(cosx+sinx)+C#.

Let #I=e^xcosxdx, and, J=inte^xsinxdx# Using IBP # ;intuvdx=uintvdx-int[(du)/dxintvdx]dx#, with, #u=cosx and, v=e^x#, we get, #I=e^xcosx-int(-sinx)e^xdx=e^xcosx+inte^xsinxdx#, i.e., #I=e^xcosx+J rArr I-J=e^xcosx.... .................(1)# Again by IBP, in #J# we get, #J=e^xsinx-inte^xcosx#, thus, #J=e^xsinx-I rArr J+I=e^xsinx.................(2)# Solving #(1) & (2)# for #I and J#, we have, #I=e^x/2(cosx+sinx)+C, and, J=e^x/2(sinx-cosx)+K# Enjoy Maths.!

Cameron Alexander · Answer

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

Going to have to use integration by parts twice.

For #u(x) and v(x)#, IBP is given by

#int uv' dx = uv - int u'vdx#

Let #u(x) = cos(x) implies u'(x) = -sin(x)#

#v'(x) = e^x implies v(x) = e^x#

#int e^xcos(x)dx = e^xcos(x) + color(red)(inte^xsin(x)dx)#

Now use IBP on the red term.

#u(x) = sin(x) implies u'(x) = cos(x)#

#v'(x) = e^x implies v(x) = e^x#

#int e^xcos(x)dx = e^xcos(x) + [e^xsin(x) - inte^xcos(x)dx]#

Group the integrals together:

#2int e^xcos(x)dx = e^x(cos(x)+sin(x)) + C#

Therefore

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

Emily Ammerman · Answer

Let #I=inte^xcosxdx#

We use,

The Rule of Integration by Parts # : intuvdx=uintvdx-int[(du)/dxintvdx]dx#.

We take, #u=cosx, and, v=e^x#.

Hence, #(du)/dx=-sinx, and, intvdx=e^x#. Therefore,

#I=e^xcosx+inte^xsinxdx=e^xcosx+J, J=inte^xsinxdx#.

To find #J#, we apply the same Rule, but, now, with #u=sinx#, &,

#v=e^x#, we get,

#J=e^xsinx-inte^xcosxdx=e^xsinx-I#.

Sub.ing this in #I#, we have,

#I=e^xcosx+e^xsinx-I#, i.e.,

#2I=e^x(cosx+sinx)#, or,

#I=e^x/2.(cosx+sinx)#.

Enjoy Maths.!

Christopher Allers · Answer

undefined To integrate e^x*cos(x), you can use integration by parts.