Integration

**Erghhh** · March 10th 2009, 01:26 PM

Integral (e^x)cosx dx

Not sure how to start?

Thanks.

**Prove It** · March 10th 2009, 01:28 PM

Originally Posted by Erghhh

Integral e^xcosx dx

Not sure how to start?

Thanks.

Take the DERIVATIVE twice, and you should find you get a nice equation which you can solve for $\int{e^x\cos{x}\,dx}$ .

Or use Integration by Parts twice.

**Erghhh** · March 10th 2009, 01:45 PM

so i got integral -2(e^x)sinx dx?

Haven't used this method of integration. Still unsure of what to do.

**Prove It** · March 10th 2009, 08:28 PM

Originally Posted by Erghhh

Integral (e^x)cosx dx

Not sure how to start?

Thanks.

$\frac{d}{dx}\left[e^x\cos{x}\right] = e^x\cos{x} - e^x\sin{x}$

So $\int{e^x\cos{x} - e^x\sin{x}\,dx} = e^x\cos{x} + C$

$\int{e^x\cos{x}\,dx} - \int{e^x\sin{x}\,dx} = e^x\cos{x} + C$ . Call this equation 1.

Now we need to find $\int{e^x\sin{x}\,dx}$ . We use the exact same method.

$\frac{d}{dx}\left[e^x\sin{x}\,dx\right] = e^x\sin{x} + e^x\cos{x}$

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

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

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

Substitute this back into equation 1.

$\int{e^x\cos{x}\,dx} - \left[e^x\sin{x} - \int{e^x\cos{x}\,dx}\right] = e^x\cos{x} + C$

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

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

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

Note, C can take on any arbitrary constant, but its actual value will change throughout the integration... I've just used the same symbol each time because I'm lazy...

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