I don't understand the result of this: $\int e^x\cos(2x) \, dx$

Question

Sorry for the simple question, I think when I find the answer I'll be like "ow how I didn't notice this..." Unfortunately I'm bad at math and i need help on this. So I don't understand the result of this: $$\int e^x\cos(2x)\,dx$$

Okay I integrated by parts twice and got this: $$e^x\cos(2x)-\left( -2e^x\sin(2x)+4\int e^x\cos(2x)\,dx \right)$$

So we found our inital integral here and so we can isolate it. But the result is: $$\int e^x\cos(2x)\,dx = \frac{2e^x\sin(2x)+e^x\cos(2)}5 +C$$

I'm having problem isolating our integral. I have no idea what happened to the 4 multiplying the integral and I have no idea where this $5$ came from.

Answer 1

See that the original integral popped out in the right side. So call your integral:

$$I=\int e^x \cos{2x} \space dx$$

So now you have that: $$I=e^x \cos{2x}+2e^x \sin{2x} -4I$$

And now you can just solve for I (you have an equation), and you have your integral (plus the constant, of course):

$$I=\frac{e^x \cos{2x}+2e^x \sin{2x} }{5}+C$$

Answer 2

Note that if we distribute the minus sign as Zach's comment suggested, we get

$$\int e^x\cos(2x)\,dx=e^x\cos(2x)+2e^x\sin(2x)-4\int e^x\cos(2x)\,dx).$$

Furthermore, if I add $4\int e^x\cos(2x)\,dx$ to both sides, I then have

$$5\int e^x\cos(2x)\,dx=e^x\cos(2x)+2e^x\sin(2x).$$ Finally, dividing both sides by $5$ achieves $$\int e^x\cos(2x)\,dx=\frac{e^x\cos(2x)+2e^x\sin(2x)}{5},$$ to which you would only need add a $+C$ to.

Answer 3

\begin{align} & \int e^x\cos(2x)\,dx \\[10pt] = {} & e^x\cos(2x)-\left( -2e^x\sin(2x)+4\int e^x\cos(2x)\,dx \right) \\[10pt] \text{Adding } & 4\int e^x \cos(2x)\,dx \text{ to both sides yields} \\[10pt] & 5\int e^x\cos(2x)\,dx = e^x\cos(2x) + 2e^2\sin(2x) + \text{constant.} \\[10pt] \text{Then } & \text{divide both sides by $5$.} \end{align}

Answer 4

Adding to @Villa's answer and now that you know how to solve it, you may replace $e^{x}$ by $e^{ax}$ and $\cos(2x)$ by $\cos(bx)$ then you can derive that:

$$\int e^{ax}\cos(bx)\,dx=\frac{e^{ax}}{a^2+b^2}\left(a\cos(bx)+b\sin(bx)\right)$$

Also,

$$\int e^{ax}\sin(bx)\,dx=\frac{e^{ax}}{a^2+b^2}\left(a\sin(bx)-b\cos(bx)\right)$$

This will save you some time.