integrate by parts e^x cos x dx?

Answer 1

f <-- integrate

You should know this: f(udv) = uv - f(vdu)

say: u = e^x, dv = cosxdx
=> du=e^x dx, v=sinx

f(e^x cos x dx) = e^x sinx - f(e^x sinxdx)

Do the same with f(e^xsinxdx)

=> f(e^x cos x dx) = e^x sinx - ( - e^x cosx + f(e^x cosxdx))

<=> f(e^x cos x dx) = 1/2(e^x sinx + e^x cosx) + C

Answer 2

oh yuck

Since e^x is derivative of e^x, we can write

A = INT e^x cos x dx = e^x cos x + INT e^x sin x dx

similarly,

B = INT e^x sin x dx = e^x sin x - INT e^x cos x dx

so we find

A = e^x cos x + B = e^x cos x + e^x sin x - A

therefore

2A = e^x (cos x + sin x)

A = INT e^x cos x dx = e^x (cos x + sin x)/2

(and as usual you may add an additive constant)

Answer 3

=e^x(sinx +cosx) /2