1. ## Integration by parts

I am half way stuck with this integral, any help would be appreciated.

Integral (p^5)ln(p) dp

u = ln p, du = 1/p
dv = p^5, v = p^6/6

Integral (p^5)ln(p) dp = p^6/6 ln(p) - Integral p^6/6p dp

Im stuck at the integral, substitution doesn't seem to work if I let w = p^6, dw = 6p^5..

2. Originally Posted by ff4930
Integral p^6/6p dp
$\int\frac{p^6}{6p}\,dp=\frac16\int p^5\,dp$

No need substitution, just apply the power rule.

3. Hello again, I stumbled with another difficult integral and need help. I didn't want to waste bandwidth so I added to this current thread.

I have integral e^2x sin3x dx

u = sin3x, du = 3cos3x
dv = e^2x, v = 1/2e^2x

integral e^2x sin3x dx = 1/2 e^2x sin3x - 3/2 integral e^2x cos3x dx

for integral e^2x cos3x dx
w = cos3x, dw = -3sin3x
dv = e^2x, v = 1/2e^2x

integral e^2x cos3x dx = 1/2e^2x cos3x + 3/2 integral e^2x sin3x dx

so combined is..

integral e^2x sin3x dx = 1/2 e^2x sin3x - 3/4 e^2x cos3x - 9/4 integral e^2x sin3x dx

since I have integral e^2x sin3x dx on both sides can I add integral e^2x sin3x dx making it

2 integral e^2x sin3x dx = 1/2 e^2x sin3x - 3/4 e^2x cos3x - 9/4

then divide both sides by 2

getting...

integral e^2x sin3x dx = [1/2 e^2x sin3x - 3/4 e^2x cos3x - 9/4]/2 + C?

4. First, set $y=3x$, then apply integration by parts.

5. Sorry, I don't quite understand what you mean by set y = 3x.

6. Originally Posted by ff4930
integral e^2x sin3x dx
$y=3x\implies dy=3\,dx$, the integral becomes

$\int e^{2x}\sin3x\,dx=\frac13\int e^{\frac23y}\sin y\,dy$

It's to simply a little bit the integrand . Now apply integration by parts.

7. It did simplified but I'm not good with variable fraction exponents.

so 1/3 integral e^(2/3y) sin y dy

u = sin y, du = cos y
dv = e^(2/3y) v = ???

unless Im not seeing something Im suppose to.
Sorry if Im not getting it right away.

8. Hello, ff4930!

A small arithmetic error in there . . . I'll do it from scratch.

We have: . $\int e^{2x}\sin3x\,dx$

By parts: . $\begin{array}{ccccccc}u & = & \sin3x & \quad & dv & = & 3\cos3x\,dx \\
du & = & 3\cos3x\,dx & \quad & v & = &\frac{1}{2}e^{2x}\end{array}$

Then: . $I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{2}\int e^{2x}\cos3x\,dx$

By parts again:\ . $\begin{array}{ccccccc}u & = & \cos3x & \quad & dv & = & e^{2x}dx \\ du & = & -3\sin3x\,dx & \quad & v & = & \frac{1}{2}e^{2x} \end{array}$

And we have: . $I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{2}\left[\frac{1}{2}e^{2x}\cos3x + \frac{3}{2}\int e^{2x}\sin3x\,dx\right]$

. . which simplifies to: . $I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{4}e^{2x}\cos3x - \frac{9}{4}\underbrace{\int e^{2x}\sin3x\,dx}_{\text{This is }I}$

So we have: . $I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{4}e^{2x}\cos3x - \frac{9}{4}I + C$

. . Then: . $\frac{13}{4}I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{4}e^{2x}\cos3x + C \;=\;\frac{1}{4}e^{2x}\left(2\sin3x - 3\cos3x\right) + C$

Multiply by $\frac{4}{13}\!:\;\;I \;=\;\frac{1}{13}\left(2\sin3x - 3\cos3x\right) + C$

. . Therefore: . $\int e^{2x}\sin3x\,dx \;=\;\frac{1}{13}\left(2\sin3x - 3\cos3x\right) + C$

I'm confuse on how you get
13/4 on the left side, also did you add 9/4 I on both sides? so shouldn't it be 2I 13/4?

$\frac{13}{4}I \;=\;\frac{1}{2}e^{2x}\sin3x - \frac{3}{4}e^{2x}\cos3x + C \;=\;\frac{1}{4}e^{2x}\left(2\sin3x - 3\cos3x\right) + C
$

10. Originally Posted by Soroban
. . Therefore: . $\int e^{2x}\sin3x\,dx \;=\;\frac{1}{13}\left(2\sin3x - 3\cos3x\right) + C$
Actually, it should be $\int e^{2x}\sin3x\,dx=\frac1{13}e^{2x}(2\sin3x-3\cos3x)+k$

In general

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

You (to ff4930) can derive this formula, of course, applyin' integration by parts.