# inverse trig integration:two methods=two diff ans HELP!!

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• April 1st 2008, 05:35 PM
i_zz_y_ill
inverse trig integration:two methods=two diff ans HELP!!
Q is whats the integral of (3)divided by(9x^2+6x+5) with no limits....

...method 1. ..............9( x^2 + (2/3)(x) + (5/9) )
=9( (x^2 + 1/3 )^2 + 4/9 )
therefore cancel 3 with 9 outside integral brackets to give 1/3 and times by
1/a i.e where a^2 = 4/9 which gives 1/2.....
.............1/2arctan 3/2(x^2 + 1/3 ) + c which gives the actual answer ....................1/2arctan(3x+1)/2 + c

method 2. which i see as perfectly ok please correct me........

(9x^2 +6x +5) simplifies to (3x + 1)^2 +4 ..........therefore taked 3 on top outside bracket and a^2 = 4.......and 3/2arctan(3x+1)/2...............

.........i dont understand why the mix up with 3/2 and 1/2 came from ....shouldn't happen..
• April 1st 2008, 05:50 PM
Jhevon
Quote:

Originally Posted by i_zz_y_ill
Q is whats the integral of (3)divided by(9x^2+6x+5) with no limits....

...method 1. ..............9( x^2 + (2/3)(x) + (5/9) )
=9( (x^2 + 1/3 )^2 + 4/9 )
therefore cancel 3 with 9 outside integral brackets to give 1/3 and times by
1/a i.e where a^2 = 4/9 which gives 1/2.....
.............1/2arctan 3/2(x^2 + 1/3 ) + c which gives the actual answer ....................1/2arctan(3x+1)/2 + c

method 2. which i see as perfectly ok please correct me........

(9x^2 +6x +5) simplifies to (3x + 1)^2 +4 ..........therefore taked 3 on top outside bracket and a^2 = 4.......and 3/2arctan(3x+1)/2...............

.........i dont understand why the mix up with 3/2 and 1/2 came from ....shouldn't happen..

you have $3 \int \frac 1{(3x + 1)^2 + 4}~dx = \frac 34 \int \frac 1{\left( \frac {3x + 1}2 \right)^2 + 1}~dx$

now make a substitution $u = \frac {3x + 1}2$
• April 1st 2008, 06:01 PM
Mathstud28
It is just
1/2arctan((3x+1)/2)+C
• April 1st 2008, 06:06 PM
i_zz_y_ill
still confused
i still dont understand . taking 3/4 out doesnt give 1/2arctan(3x+1)/2 ,,,,

this is wher i got the method from integral of 4/(x^2 -2x = 3)
=4/root2 arctan (x-1/root2) + c
• April 1st 2008, 06:08 PM
Mathstud28
Because
int[u'(x)/(a+(u(x))²)]=1/arctan(u(x)/a)+C
• April 1st 2008, 06:09 PM
Jhevon
Quote:

Originally Posted by i_zz_y_ill
i still dont understand . taking 3/4 out doesnt give 1/2arctan(3x+1)/2 ,,,,

this is wher i got the method from integral of 4/(x^2 -2x = 3)
=4/root2 arctan (x-1/root2) + c

that is why i told you to use substitution. do you know how to do integration by substitution?
• April 1st 2008, 06:14 PM
i_zz_y_ill
still confused....
k erm... i was suing (int) 1/a^2 + (f(x))^2 = 1/a arctan x/a + c. Erm then i just took the 3 out??????
• April 1st 2008, 06:16 PM
Mathstud28
Except
that it isnt always x and you need to have the derivative of the quantity inside the ()² on the bottom in this case the derivative of 3x+1=3 which you have on teh top...therfore you can just integrate it
• April 1st 2008, 06:19 PM
h2osprey
There’s nothing wrong with your second method, save for the last part: you should get 3 [tan^-1 (3x+1 / 2)] . 2 . 1/3 , which will give you 1/3 [tan^-1 (3x+1 / 2)]
• April 1st 2008, 06:20 PM
Mathstud28
Just....
Just study int[u'(x)/(1+u(x)²)]=1/a*arctan(u(x)/a)+C
• April 1st 2008, 06:22 PM
Jhevon
Picking up where we left off. yes, you can use the formula in your text book, but obviously it won't work if you don't know how to use it.

we wish to find $\frac 34 \int \frac 1{\left( \frac {3x + 1}2 \right)^2 + 1}~dx$

Let $u = \frac {3x + 1}2$

$\Rightarrow du = \frac 32~dx$

$\Rightarrow \frac 23 ~du = dx$

So our integral becomes:

$\frac 34 \cdot \frac 23 \int \frac 1{u^2 + 1}~du = \frac 12 \arctan u + C$

back-substitute the expresion for $u$, we get

$\frac 12 \arctan \frac {3x + 1}2 + C$
• April 1st 2008, 06:24 PM
h2osprey
Quote:

Originally Posted by h2osprey
There’s nothing wrong with your second method, save for the last part: you should get 3 [tan^-1 (3x+1 / 2)] . 2 . 1/3 , which will give you 1/3 [tan^-1 (3x+1 / 2)]

Sorry, I meant

There’s nothing wrong with your second method, save for the last part: you should get 3 [tan^-1 (3x+1 / 2)] . 1/2 . 1/3 , which will give you 1/2[tan^-1 (3x+1 / 2)][/QUOTE]
• April 1st 2008, 06:24 PM
i_zz_y_ill
still not sure!
k so is i have 3/( (3x + 1)^2 +4) integrated gives 1/arctan(3x + 1)/4.......................this isnt 1/2arctan(3x+1)/2 which is ans is thats what your saying?
• April 1st 2008, 06:30 PM
h2osprey
No, he's saying that the given answer is correct.. in fact you probably just forgot to divide by 3 in the second method. Due to the fact that the coefficient of x in (3x+1)^2 is 3, you have to take that into account and divide accordingly when evaluating the integral. (Sort of like reverse differentiation, if you have 3x you multiply by 3, hence when you integrate you must do the opposite).
• April 1st 2008, 06:30 PM
i_zz_y_ill
cheers but:p...
when u take 4 out side with the 3/4,,,,,,,,(this is jhevs method), doesnt it become
3/4 (integral) 1/((3x+1)^2)/4 + 1

not 3/3 (integral) 1/((3x+1)^2)/2 +1
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