Integration - First Steps

**greenstupor** · February 7th 2010, 04:24 PM

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.

Thanks in advance!

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

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

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

**pickslides** · February 7th 2010, 04:30 PM

2. $\int\frac{x^3}{x^2-2x+3}~dx= \int x+2+\frac{x-6}{x^2-2x+3}~dx$

**General** · February 7th 2010, 05:16 PM

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.

Thanks in advance!

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

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

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

1. Let $\sqrt{7}t=5sin(\theta)$ .
3. By parts, $u=cos(4x)$ and $dv= e^{3x}dx$ .

**greenstupor** · February 7th 2010, 05:25 PM

Thanks for all your help!

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

$uv - \int vdu$

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

What should I do?

**greenstupor** · February 7th 2010, 06:24 PM

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

**General** · February 7th 2010, 10:43 PM

Originally Posted by greenstupor

Thanks for all your help!

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

$uv - \int vdu$

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

What should I do?

Do by parts again.

**General** · February 7th 2010, 10:43 PM

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?

**HallsofIvy** · February 8th 2010, 04:15 AM

Originally Posted by greenstupor

Thanks for all your help!

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

$uv - \int vdu$

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

What should I do?

If you integrate by parts again, you will get something like
$\int e^{3x}sin(4x)dx= A+ B\int e^{3x}sin(4x) dx$ .

Solve that equation for $\int e^{3x}sin(4x) dx$

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