\int e^{\alpha x} \cos \beta x \mathrm{d}x , x \epsilon R , \alpha^2 + \beta^2 \neq 0

Hint: use integration by parts, two times. First time set u = e^(ax) and dv = cos(bx)dx. I'll let you figure out the rest!

Hint: use integration by parts, two times. First time set u = e^(ax) and dv = cos(bx)dx. I'll let you figure out the rest!

Here is what I've got:

\int e^{\alpha x} \cos \beta x \mathrm {d} x = \int \(\frac{1}{\alpha} e^{\alpha x}\)' \cos \beta x \mathrm{d}x = \frac{1}{\alpha} e^{\alpha x} \cos \beta x + \frac{1}{\alpha} \int e^{\alpha x} \sin \beta x \mathrm{d}x = \frac{1}{\alpha} e^{\alpha x} \cos \beta x + \frac{1}{\alpha} \int \(\frac{1}{\alpha} e^{\alpha x}\)' \sin \beta x \mathrm{d}x = \frac{1}{\alpha} e^{\alpha x} \cos \beta x + \frac{1}{\alpha ^ 2} e^{\alpha x} \sin \beta x - \frac{1}{\alpha ^2} \int e^{\alpha x} \cos \beta x \mathrm{d}x

In conclusion: \(1+ \frac{1}{\alpha ^2}\) \int e^{\alpha x} \cos \beta x \mathrm{d}x = \frac{1}{\alpha} e^{\alpha x} \(\cos \beta x + \frac{1}{\alpha} \sin \beta x \)

But what get here is extremely different from the correct answer:
\int e^{\alpha x} \cos \beta x \mathrm {d} x = \frac{e^{\alpha x}}{\alpha ^2 + \beta ^2} (\alpha \sin \beta x + \beta \cos \beta x)

So where is the mistake?

So where is the mistake??

The mistake is after the second equals sign, you should have:


\frac{1}{\alpha} e^{\alpha x} \cos \beta x + \frac{\beta}{\alpha} \int e^{\alpha x} \sin \beta x \mathrm{d}x


Note the beta before the integral, which is needed because if u = \cos(\beta x) then du = \beta \sin(\beta x). Same mistake with the second integration by parts. So the "in conclusion" line should be:


(1 + \frac {\beta^2}{\alpha ^2}) \int e^{\alpha x} \sin \beta x dx =( \frac 1 a ) e^{ax} \cos \beta x +( \frac \beta {\alpha ^2} ) e^{ax} \sin \beta x

The mistake is after the second equals sign, you should have:


\frac{1}{\alpha} e^{\alpha x} \cos \beta x + \frac{\beta}{\alpha} \int e^{\alpha x} \sin \beta x \mathrm{d}x


Note the beta before the integral, which is needed because if u = \cos(\beta x) then du = \beta \sin(\beta x). Same mistake with the second integration by parts. So the "in conclusion" line should be:


(1 + \frac {\beta^2}{\alpha ^2}) \int e^{\alpha x} \sin \beta x dx =( \frac 1 a ) e^{ax} \cos \beta x +( \frac \beta {\alpha ^2} ) e^{ax} \sin \beta x


Thank you!