# Thread: Integration - First Steps

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

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

3. 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 $\displaystyle dv= e^{3x}dx$.

4. Thanks for all your help!

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?

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

6. 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.

7. 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?

8. 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?
If you integrate by parts again, you will get something like
$\displaystyle \int e^{3x}sin(4x)dx= A+ B\int e^{3x}sin(4x) dx$.

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