1. ## Integration

These few questions have been bugging me for hours, help would be appreciated

2. ## Re: Integration

1. What's the lower bound? Use the goniometric substitution $\displaystyle t=\tan(u)$
2. Use integration by parts.

3. ## Re: Integration

Originally Posted by Siron
1. What's the lower bound? Use the goniometric substitution $\displaystyle t=\tan(u)$
2. Use integration by parts.
Thank you How do you know when to use t = tanu?

Also would you be able to give me an indication of what to do for this one:

Integrate 9t/(9t^2 + 6t +5)

4. ## Re: Integration

Originally Posted by qwerty31
Thank you How do you know when to use t = tanu?
Because with that subsitution you can 'eliminate' the root form which is easier to work with. (Have you solved the integral?)

Originally Posted by qwerty31
Also would you be able to give me an indication of what to do for this one:
Integrate 9t/(9t^2 + 6t +5)
$\displaystyle \int \frac{9t}{9t^2+6t+5}dt = \frac{1}{2} \int \frac{(18t+6)-6}{9t^2+6t+5}dt = \frac{1}{2} \left(\int \frac{18t+6}{9t^2+6t+5}dt - \int \frac{6}{9t^2+6t+5}dt\right)$

I guess you can take it from here. To solve $\displaystyle \int \frac{6}{9t^2+6t+5}dt$ write $\displaystyle 9t^2+6t+5=(3t+1)^2+4$ (because $\displaystyle D<0$)

5. ## Re: Integration

$\displaystyle sin^2u+ cos^2u= 1$ so, dividing both sides by $\displaystyle cos^2 u$, $\displaystyle tan^2u+ 1= sec^2u$. Because of that the substitution $\displaystyle x= tan u$ changes $\displaystyle \sqrt{1+ x^2}$ to $\displaystyle \sqrt{1+ tan^2 u}= \sqrt{sec^2 u}= sec u$. Generally speaking, any time you have a square root of a quadratic, you should consider a trig substitution. (Some people prefer hyperbolic functions which satisfy similar identities.)

As for your new question, first complete the square: $\displaystyle 9t^2+ 6t+ 5= 9(t^2+ (2/3)t+ 1/9- 1/9)+ 5)= 9(t+ 1/3)^2+ 4$.
Now the substitution x= t+ 1/3 changes to integrand to $\displaystyle \frac{9u- 3}{9u^2+ 4}= \frac{u}{u^2+ (2/3)^2}+ (4/9)\frac{1}{(3u/2)^2+ 1}$