Question

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

Answer 1

`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`

Answer 2

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.!

Answer 3

`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.!