Integration

**qwerty31** · April 13th 2012, 12:18 PM

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

**Siron** · April 13th 2012, 12:35 PM

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

**qwerty31** · April 14th 2012, 03:00 AM

Originally Posted by Siron

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

**Siron** · April 14th 2012, 05:39 AM

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)

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

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

**HallsofIvy** · April 14th 2012, 05:47 AM

$sin^2u+ cos^2u= 1$ so, dividing both sides by $cos^2 u$ , $tan^2u+ 1= sec^2u$ . Because of that the substitution $x= tan u$ changes $\sqrt{1+ x^2}$ to $\sqrt{1+ tan^2 u}= \sqrt{sec^2 u}= sec u$ . 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: $9t^2+ 6t+ 5= 9(t^2+ (2/3)t+ 1/9- 1/9)+ 5)= 9(t+ 1/3)^2+ 4$ .
Now the substitution x= t+ 1/3 changes to integrand to $\frac{9u- 3}{9u^2+ 4}= \frac{u}{u^2+ (2/3)^2}+ (4/9)\frac{1}{(3u/2)^2+ 1}$

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