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dontletmego · Nov 1, 2012, 10:21 AM

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

ebaines · Nov 1, 2012, 11:04 AM

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!

dontletmego · Nov 1, 2012, 11:36 AM

Originally Posted by ebaines

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?

ebaines · Nov 1, 2012, 11:58 AM

Originally Posted by dontletmego

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 $

dontletmego · Nov 2, 2012, 11:29 AM

Originally Posted by ebaines

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!