Integral Solution Calculation

Question

Question:

Using various techniques of integration: Determine the exact value of $\int _0^1\:e^{3x}cos\left(2x\right)\:dx$ .

Attempt:

Integration by Parts step:

Let $u = e^{3x}$ and $dv = \cos(2x) dx$.

Then, $du = 3e^{3x}dx$ and $v = \frac{1}{2}\sin(2x)$.

Using integration by parts:

$$I = uv\Big|_0^1 - \int v du$$

$$I = \frac{1}{2}e^{3x}\sin(2x)\Big|_0^1 - \int \frac{1}{2}\sin(2x)(3e^{3x}) dx$$

$$I = \frac{1}{2}e^3\sin(2) - \frac{3}{2}\int e^{3x}\sin(2x) dx$$

Now, let's use integration by parts again:

Let $u = e^{3x}$ and $dv = \sin(2x) dx$.

Then, $du = 3e^{3x}dx$ and $v = -\frac{1}{2}\cos(2x)$.

$$-\frac{3}{2}\int e^{3x}\sin(2x) dx = -\frac{3}{2}\left[-\frac{1}{2}e^{3x}\cos(2x)\Big|_0^1 - \int -\frac{1}{2}\cos(2x)(3e^{3x}) dx\right]$$

$$-\frac{3}{2}\int e^{3x}\sin(2x) dx = \frac{3}{4}e^{3x}\cos(2x)\Big|_0^1 + \frac{9}{4}\int e^{3x}\cos(2x) dx$$

So, we have:

$$I = \frac{1}{2}e^3\sin(2) - \frac{3}{4}e^3\cos(2) + \frac{3}{4} - \frac{9}{4}I$$

Now, solve for $I$:

$$I + \frac{9}{4}I = \frac{1}{2}e^3\sin(2) - \frac{3}{4}e^3\cos(2) + \frac{3}{4}$$

$$\frac{13}{4}I = \frac{1}{2}e^3\sin(2) - \frac{3}{4}e^3\cos(2) + \frac{3}{4}$$

$$I = \frac{4}{13}\left[\frac{1}{2}e^3\sin(2) - \frac{3}{4}e^3\cos(2) + \frac{3}{4}\right]$$

The correct answer for the integral is:

$$I = \frac{2e^3\sin(2) - 3e^3\cos(2) + 3}{13}$$

Is this correct ?

Answer 1

using integration by parts twice $$I＝\int_{0}^{1}e^{3x}cos(2x)dx＝\left({\frac {e^{3x}}{3}cos(2x)}\right)^1_0＋\textstyle\frac {2}{3}\int_{0}^{1}e^{3x}sin(2x)dx\\~\\＝\frac {e^{3}}{3}cos(2)-\frac {1}{3}＋\frac {2}{3}\left({\left({\frac {e^{3x}}{3}sin(2x)}\right)^1_0－\frac {2}{3}\int_{0}^{1}e^{3x}cos(2x)dx}\right)\\~\\＝\frac {e^{3}}{3}cos(2)-\frac {1}{3}＋\frac {2}{3}(\frac {e^{3}}{3}sin(2))－\frac {4}{9}I\\~\\\implies\;I＋\frac {4}{9}I＝\frac {e^{3}}{3}cos(2)-\frac {1}{3}＋\frac {2e^{3}}{9}sin(2)\\~\\I＝\frac {9}{13}(\frac {e^{3}}{3}cos(2)-\frac {1}{3}＋\frac {2e^{3}}{9}sin(2))\\~\\$$ And therefore:$$I＝\frac {3e^{3}cos(2)-3＋2e^{3}sin(2)}{13}$$

Answer 2

So faster without integration by parts at all $$I=\int \:e^{3x}\cos\left(2x\right)\:dx=\Re\left(\int\:e^{(3+2i)x}\:dx \right)=\Re\left(\frac{3-2i}{13}\, e^{(3+2 i) x}\right)$$ $$I=\int \:e^{3x}\cos\left(2x\right)\:dx=\frac{1}{13} e^{3 x}\, (2 \sin (2 x)+3 \cos (2 x))$$

Answer 3

Let $f=e^{3x}\cos 2x.$ Let $g=e^{3x}\sin 2x.$

Then $f'=3f-2g$ and $g'=3g+2f.$

So for constants $a,b$ we have

$(af+bg)'-f=(3a+2b-1)f+(-2a+3b)g .$

This is $0$ for all $x$ if (and only if) $0=3a+2b-1=-2a+3b,$ that is, $a=3/13$ and $b=2/13.$

So $\int f=(3f+2g)/13.$

This method generalizes for integrating $e^{px}\cos qx$ or $e^{px}\sin qx$ with constants $p,q.$ If you examine the formulas in the Answer by @Claude Leibovici, you can see why $\int f$ should be $af+bg$ for some constants $a,b.$

Addendum: Your approach is valid but it is easy to make errors when calculating this way. It is usually safer when doing repeated integration by parts to first try it with the $indefinite$ integrals.