integral of e^2x * sinx

**mottin** · March 31st 2010, 05:14 PM

i gotta integrate the following:

e^2x * sinx

i got this as my answer:

(4/5)(-e^2x*cosx + .5e^2x * sinx)

is it right?

**pickslides** · March 31st 2010, 05:18 PM

Is done by parts twice.

Wolfram Mathematica Online Integrator

**Prove It** · March 31st 2010, 05:39 PM

An alternative method is to take the derivative twice...

$\frac{d}{dx}(e^{2x}\sin{x}) = e^{2x}\cos{x} + 2e^{2x}\sin{x}$ .

Therefore $\int{e^{2x}\cos{x} + 2e^{2x}\sin{x}\,dx} = e^{2x}\sin{x}$

$\int{e^{2x}\cos{x}\,dx} + 2\int{e^{2x}\sin{x}\,dx} = e^{2x}\sin{x}$

$\int{e^{2x}\sin{x}\,dx} = \frac{1}{2}e^{2x}\sin{x} - \frac{1}{2}\int{e^{2x}\,\cos{x}\,dx}$ .

Call this equation 1. Notice that now we have another integral to find... so take the derivative again...

$\frac{d}{dx}(e^{2x}\cos{x}) = 2e^{2x}\cos{x} - e^{2x}\sin{x}$ .

Therefore $\int{2e^{2x}\cos{x} - e^{2x}\sin{x}\,dx} = e^{2x}\cos{x}$

$2\int{e^{2x}\cos{x}\,dx} - \int{e^{2x}\sin{x}\,dx} = e^{2x}\cos{x}$

$\int{e^{2x}\cos{x}\,dx} = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{2}\int{e^{2x}\sin{x}\,dx}$ .

Substitute this back into equation 1:

$\int{e^{2x}\sin{x}\,dx} = \frac{1}{2}e^{2x}\sin{x} - \frac{1}{2}\left(\frac{1}{2}e^{2x}\cos{x} + \frac{1}{2}\int{e^{2x}\sin{x}\,dx}\right)$

$\int{e^{2x}\sin{x}\,dx} = \frac{1}{2}e^{2x}\sin{x} - \frac{1}{4}e^{2x}\cos{x} - \frac{1}{4}\int{e^{2x}\sin{x}\,dx}$

$\frac{5}{4}\int{e^{2x}\sin{x}\,dx} = \frac{1}{2}e^{2x}\sin{x} - \frac{1}{4}e^{2x}\cos{x}$

$\int{e^{2x}\sin{x}\,dx} = \frac{2}{5}e^{2x}\sin{x} - \frac{1}{5}e^{2x}\cos{x} + C$ .

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