Integration - First Steps

[greenstupor]

Hey everybody,

My class is learning integration by parts and integration by substitution right now, and I'm having some trouble with them.

Could anybody please help me with these 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!

Antiderivatives of:
1. [1/(25 - 7t^2)]dt

• I don't even know the first step to take with this.[/*:m:2i47gmmn]

2. [(x^3)/(x^2 - 2x + 3)]dx

• I could convert this to the antiderivatives of: x + 2 + (x - 6)/(x^2 - 2x + 3), but is that the only way to do it? Or can I just do integration by parts or substitution from the very beginning?[/*:m:2i47gmmn]

3. [e^(3x)cos(4x)]dx

• I've tried making u = e^(3x) and dv = cos(4x)dx, but then I keep going around in circles when I try to do integration by parts. [/*:m:2i47gmmn]

 

[NRS]

I don't get to integration by parts for a couple of weeks or so, but I would say for the first one, would it work if you were to substitute u for 7t?

 

[greenstupor]

I don't get to integration by parts for a couple of weeks or so, but I would say for the first one, would it work if you were to substitute u for 7t?

I'm not sure, but I don't think you can substitute u for 7t because the problem has 7(t^2).

 

[NRS]

sorry 'bout that, you could sub it for just t though.

 

[BigGlenntheHeavy]

\(\displaystyle I'll \ do \ the \ first \ one \ for \ you, \ to \ get \ you \ started.\)

\(\displaystyle \int\frac{dt}{25-7t^{2}} \ = \ \int\frac{dt}{5^{2}-7t^{2}}\)

\(\displaystyle Let \ u \ = \ \sqrt7t, \ then \ du \ = \ \sqrt7dt\)

\(\displaystyle Hence, \ \int\frac{dt}{5^{2}-7t^{2}} \ =\frac{1}{\sqrt7} \ \int\frac{du}{5^{2}-u^{2}}\)

\(\displaystyle = \ \frac{1}{10\sqrt7}\int\frac{du}{u+5}-\frac{1}{10\sqrt7}\int\frac{du}{u-5}, \ Partial \ fractions.\)

\(\displaystyle = \ \frac{1}{10\sqrt7}ln|u+5|-\frac{1}{10\sqrt7}ln|u-5| \ +C\)

\(\displaystyle = \ \frac{1}{10\sqrt7}ln\bigg|\frac{|u+5|}{|u-5|}\bigg|+C \ = \ \frac{\sqrt7}{70}ln\bigg|\frac{|\sqrt7t+5|}{|\sqrt7t-5|}\bigg|+C\)

\(\displaystyle Note: \ A \ good \ check \ (and \ a \ good \ exercise \ for \ you) \ is \ to \ take \ the \ derivative \ of \ your \ final\)

\(\displaystyle answer. \ If \ it \ is \ correct, \ then \ you \ should \ have \ what \ you \ originally \ started \ with, \ to \ wit:\)

\(\displaystyle D_t \ \bigg[\frac{\sqrt7}{70}ln\bigg|\frac{|\sqrt7t+5|}{|\sqrt7t-5|}\bigg|+C\bigg] \ = \ \frac{1}{25-7t^{2}}\)

 

[BigGlenntheHeavy]

\(\displaystyle Hint \ for \ the \ second \ one, \ to \ wit:\)

\(\displaystyle \int\frac{x^{3}}{x^{2}-2x+3}dx \ = \ \int\bigg[x+2+\frac{x}{(x-1)^{2}+2}-\frac{6}{(x-1)^{2}+2}\bigg]dx\)

 

[Aladdin]

\(\displaystyle Hint \ for \ the \ second \ one, \ to \ wit:\)

\(\displaystyle \int\frac{x^{3}}{x^{2}-2x+3}dx \ = \ \int\bigg[x+2+\frac{x}{(x-1)^{2}+2}-\frac{6}{(x-1)^{2}+2}\bigg]dx\)

Classical devision , right Glenn .

 

[BigGlenntheHeavy]

Aladdin, division is find, however I used partial fractions, basically, I guess you could call them the same thing (some times).

\(\displaystyle \frac{x^{3}}{x^{2}-2x+3} \ = \ x+2+\frac{x-6}{x^{2}-2x+3} \ = \ x+2+\frac{x}{x^{2}-2x+3}-\frac{6}{x^{2}-2x+3}\)

\(\displaystyle = \ x+2+\frac{x}{(x-1)^{2}+2}-\frac{6}{(x-1)^{2}+2}, \ by \ division \ only. \ Same \ results \ with \ partial \ fractions.\)

 

[galactus]

3. [e^(3x)cos(4x)]dx

• I've tried making u = e^(3x) and dv = cos(4x)dx, but then I keep going around in circles when I try to do integration by parts. [/*:m:3is2zik5]

You say you're going in circles. That is because when you do parts twice you end up with \(\displaystyle \int e^{3x}cos(4x)dx\) again on the right side. Correct?. Add it to both sides. See what I mean?.

Let \(\displaystyle u=cos(4x), \;\ dv=e^{3x}dx, \;\ du=-4sin(4x)dx, \;\ v=\frac{1}{3}e^{3x}\)

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

Do it again:

\(\displaystyle u=sin(4x), \;\ dv=e^{3x}dx, \;\ du=4cos(4x)dx, \;\ v=\frac{1}{3}e^{3x}\)

\(\displaystyle \int e^{3x}cos(4x)dx=\frac{1}{3}e^{3x}cos(4x)+\frac{4}{9}e^{3x}sin(4x)-\frac{16}{9}\int e^{3x}cos(4x)dx\)

See now what to do?. Add the \(\displaystyle \frac{16}{9}\int e^{3x}cos(4x)dx\) to both sides:

\(\displaystyle \frac{25}{9}\int e^{3x}cos(4x)dx=\frac{1}{3}e^{3x}cos(4x)dx+\frac{4}{9}e^{3x}sin(4x)\)

Multiply through by 9/25:

\(\displaystyle \int e^{3x}cos(4x)dx=\frac{3e^{3x}cos(4x)}{25}+\frac{4e^{3x}sin(4x)}{25}\)

See now how to do it?. Rookies at parts usually do not see to do this at first. But, you can.