Question

Hi guys, can you help me in this one? Use integration by parts twice to evaluate ∫e^6t cos(7t)dt :

Answer 1

I bet that is e^(6t)?

Answer 2

What did you get in your first round of integration by parts?

Answer 3

yes , it is e^(6t)

Answer 4

Let the original integral be I.
When you do integration by parts twice you will get the original integral again on the right hand side. Replace that with I, bring it to the LHS and solve for I.

Answer 5

Another tip: keep using the same (or similar) \(u\) and \(dv\) expressions. If you switch from, say \(u_1=e^{6t}\) to \(u_2=\text{(some trig)}(7t)\), you'll end up back where you started.

Answer 6

\[
I = \int e^{6t}\cos(7t)dt \\
u = \cos(7t); ~~du = -7\sin(7t)dt \\
dv = e^{6t}dt; ~~ v = \frac 16e^{6t} \\
\text{ } \\

I = \frac 16e^{6t}\cos(7t) + \frac 76\int e^{6t}\sin(7t)dt \\
u = \sin(7t); ~~du = 7\cos(7t)dt \\
dv = e^{6t}dt; ~~ v = \frac 16e^{6t} \\
\text{ } \\

I = \frac 16e^{6t}\cos(7t) + \frac 76 * \frac 16 e^{6t}\sin(7t) - \frac 76 *
\frac 76 \int e^{6t}\cos(7t)dt \\
I = \frac 16e^{6t}\cos(7t) + \frac {7}{36} e^{6t}\sin(7t) - \frac {49}{36} I \\
\text{ } \\

36I = 6e^{6t}\cos(7t) + 7 e^{6t}\sin(7t) - 49 I \\
36I + 49I = 6e^{6t}\cos(7t) + 7 e^{6t}\sin(7t) = e^{6t}\{6\cos(7t) + 7\sin(7t) \}\\
85I = e^{6t}\{6\cos(7t) + 7\sin(7t) \}\\
I = \frac{e^{6t}}{85}\{6\cos(7t) + 7\sin(7t) \} + C
\]

Answer 7

Thank you so much guys ^_^

Answer 8

You are welcome.