# I'm halfway stuck solving an Integral derivation.

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.

Page 1

Page 2

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