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

Printable View

- Apr 13th 2012, 12:18 PMqwerty31Integration
These few questions have been bugging me for hours, help would be appreciated

- Apr 13th 2012, 12:35 PMSironRe: Integration
1. What's the lower bound? Use the goniometric substitution $\displaystyle t=\tan(u)$

2. Use integration by parts. - Apr 14th 2012, 03:00 AMqwerty31Re: Integration
- Apr 14th 2012, 05:39 AMSironRe: Integration
Because with that subsitution you can 'eliminate' the root form which is easier to work with. (Have you solved the integral?)

$\displaystyle \int \frac{9t}{9t^2+6t+5}dt = \frac{1}{2} \int \frac{(18t+6)-6}{9t^2+6t+5}dt = \frac{1}{2} \left(\int \frac{18t+6}{9t^2+6t+5}dt - \int \frac{6}{9t^2+6t+5}dt\right)$

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 $\displaystyle D<0$) - Apr 14th 2012, 05:47 AMHallsofIvyRe: 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 $\displaystyle \frac{9u- 3}{9u^2+ 4}= \frac{u}{u^2+ (2/3)^2}+ (4/9)\frac{1}{(3u/2)^2+ 1}$