# Integration - First Steps

• Feb 7th 2010, 04:24 PM
greenstupor
Integration - First Steps
Could anyone please help me with these integration problems? You don't have to give me the answers, but if you could point me in the direction to get started, I'd really appreciate it.

1. $\displaystyle \int\frac{1}{\sqrt{25-7t^2}}~dt$

2. $\displaystyle \int\frac{x^3}{x^2-2x+3}~dx$

3. $\displaystyle \int e^{3x}cos{4x}~dx$
• Feb 7th 2010, 04:30 PM
pickslides
2. $\displaystyle \int\frac{x^3}{x^2-2x+3}~dx= \int x+2+\frac{x-6}{x^2-2x+3}~dx$
• Feb 7th 2010, 05:16 PM
General
Quote:

Originally Posted by greenstupor
Could anyone please help me with these integration problems? You don't have to give me the answers, but if you could point me in the direction to get started, I'd really appreciate it.

1. $\displaystyle \int\frac{1}{\sqrt{25-7t^2}}~dt$

2. $\displaystyle \int\frac{x^3}{x^2-2x+3}~dx$

3. $\displaystyle \int e^{3x}cos{4x}~dx$

1. Let $\displaystyle \sqrt{7}t=5sin(\theta)$.
3. By parts, $\displaystyle u=cos(4x)$ and $\displaystyle dv= e^{3x}dx$.
• Feb 7th 2010, 05:25 PM
greenstupor

For #3, I've tried doing that, but I keep going around in circles:

$\displaystyle uv - \int vdu$

$\displaystyle \frac{1}{3}cos(4x)e^{3x} - \frac{-4}{3}\int e^{3x}sin(4x)~dx$

What should I do? :(
• Feb 7th 2010, 06:24 PM
greenstupor
Oh, and I'm also not sure what you mean by your tip on #1. Sorry. I'm trying my best on this stuff. :(
• Feb 7th 2010, 10:43 PM
General
Quote:

Originally Posted by greenstupor

For #3, I've tried doing that, but I keep going around in circles:

$\displaystyle uv - \int vdu$

$\displaystyle \frac{1}{3}cos(4x)e^{3x} - \frac{-4}{3}\int e^{3x}sin(4x)~dx$

What should I do? :(

Do by parts again.
• Feb 7th 2010, 10:43 PM
General
Quote:

Originally Posted by greenstupor
Oh, and I'm also not sure what you mean by your tip on #1. Sorry. I'm trying my best on this stuff. :(

Did you apply this substitution on the integral?
Do you the trigonometric substitutions?
• Feb 8th 2010, 04:15 AM
HallsofIvy
Quote:

Originally Posted by greenstupor
$\displaystyle uv - \int vdu$
$\displaystyle \frac{1}{3}cos(4x)e^{3x} - \frac{-4}{3}\int e^{3x}sin(4x)~dx$
$\displaystyle \int e^{3x}sin(4x)dx= A+ B\int e^{3x}sin(4x) dx$.
Solve that equation for $\displaystyle \int e^{3x}sin(4x) dx$