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

3 Answers
Aug 13, 2016

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

Explanation:

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

Aug 13, 2016

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.!

Aug 13, 2016

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

Explanation:

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.!