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I apologize beforehand for my handwriting. I hope it is legible. I worked the problem on a piece of paper. Took a picture and attached it to this query.

I wrote the problem starting with Page 1 and then continuing on Page 2. I know what the answer should be. But I'm stuck halfway through not able to get to the answer.

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Let $I = \int e^{-\alpha t}\cos(\omega t)dt$

Let $u = e^{-\alpha t} $ and $dv = \cos(\omega t)dt$

$du = -\alpha e^{-\alpha t}dt$ and $v = \frac{\sin\omega t}{\omega}$

So, $I = uv - \int vdu$

$I = e^{-\alpha t} \frac{\sin\omega t}{\omega} - \int \frac{\sin\omega t}{\omega}(-\alpha e^{-\alpha t})dt$

$I = e^{-\alpha t} \frac{\sin\omega t}{\omega} + \frac{\alpha }{\omega} \int \sin\omega t(e^{-\alpha t})dt$

Again let $u = e^{-\alpha t}$ and $dv = \sin\omega tdt$

$du =-\alpha e^{-\alpha t} dt$ and $v = \frac{-1}{\omega} \cos\omega t$

So, $\frac{\alpha }{\omega} \int \sin\omega t(e^{-\alpha t})dt = \frac{\alpha }{\omega}\big[ \frac{-1}{\omega} \cos\omega t e^{-\alpha t} - \int \frac{-1}{\omega} \cos\omega t (-\alpha) e^{-\alpha t} dt\big] +c_1$

$\frac{\alpha }{\omega} \int \sin\omega t(e^{-\alpha t})dt = \frac{-\alpha}{\omega ^2}\cos\omega t e^{-\alpha t} - \frac{\alpha^2}{\omega^2}I + c_1$

So,

$I = e^{-\alpha t} \frac{\sin\omega t}{\omega} - \frac{\alpha}{\omega ^2}\cos\omega t e^{-\alpha t} - \frac{\alpha^2}{\omega^2}I +c_1$

$I + \frac{\alpha^2}{\omega^2}I = \frac{e^{-\alpha t}}{\omega^2}\bigg[\omega \sin \omega t - \alpha \cos \omega t\bigg] $

$I\frac{(\alpha^2 + \omega^2)}{\omega^2} = \frac{e^{-\alpha t}}{\omega^2}\bigg[\omega \sin \omega t - \alpha \cos \omega t\bigg] +c_1 $

$$I = \frac{e^{-\alpha t}}{\alpha^2+\omega^2}\bigg[\omega \sin \omega t - \alpha \cos \omega t\bigg] + C$$

Now multiply $\alpha$ and substitute the limits. Ignore the constant as it is a definite integral

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