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Integrals giving inverse trig.

paulmdrdo

Active member
May 13, 2013
386
again, i need some help here guys.


$\displaystyle\int\frac{3x-1}{2x^2+2x+3}dx$

=$\displaystyle\int\frac{3x-1}{2\left[\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\right]}dx$

$\displaystyle a=\frac{\sqrt{5}}{2}$; $\displaystyle u=x+\frac{1}{2}$; $\displaystyle du=dx$; $\displaystyle x=u-\frac{1}{2}$

=$\displaystyle\frac{1}{2}\int3\frac{\left(u-\frac{1}{2}\right)}{a^2+u^2}du$;

=$\displaystyle\frac{1}{2}\int\frac{\left(3u-\frac{3}{2}\right)}{a^2+u^2}du$

=$\displaystyle\frac{1}{2}\left(\int\frac{3u}{a^2+u^2}du-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

$\displaystyle v=a^2+u^2$; $\displaystyle dv=2udu$; $\displaystyle du=\frac{1}{2u}dv$

=$\displaystyle\frac{1}{2}\left(3\int\frac{udv}{v\,2u}dv-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

=$\displaystyle\frac{1}{2}\left(\frac{3}{2}\int \frac{dv}{v}-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

=$\displaystyle\frac{1}{2}\left(\frac{3}{2}ln(a^2+u^2)-\frac{3}{2}\frac{2}{\sqrt{5}}\arctan\frac{2x+1}{ \sqrt{5}}\right)$

=$\displaystyle\frac{3}{4}ln\left(x+\frac{\sqrt{5}+1}{2}\right)-\frac{3}{2\sqrt{5}}\arctan\frac{2x+1}{ \sqrt{5}}+C$ ---> FINAL ANSWER.

would you be so generous to check if my solution is correct?

thanks!

the answer in my book is in this form,

$\displaystyle\frac{3}{4}ln(2x^2+2x+3)-\frac{\sqrt{5}}{2}\arctan\frac{2x+1}{\sqrt{5}}+C$

is my answer equivalent to this one?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Re: integrals giving inverse trig.

Hmm. I have several thoughts on this.
  1. You can check your answer by differentiating, and seeing if you get back to the original integrand. What does that give you?
  2. My HP 50g calculator gives me
    $$\int=- \frac{ \sqrt{5}}{2} \arctan \left( \frac{ \sqrt{5}+2 \sqrt{5} x}{5} \right)+ \frac{3}{4} \ln |2x^{2}+2x+3|.$$
  3. You can see for yourself what W/A gives.
  4. In looking over your work, I think your first substitution might have gone awry. I wouldn't do more than one substitution at once.
 

paulmdrdo

Active member
May 13, 2013
386
Re: integrals giving inverse trig.

what's wrong with my first substitution? please explain.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: integrals giving inverse trig.

You have $3x-1$ if we use the substitution \(\displaystyle x=u-\frac{1}{2}\)

we get $3\left(u-\frac{1}{2} \right)-1=3u-\frac{3}{2}-1=3u-\frac{5}{2}$

You missed the \(\displaystyle -1\) in \(\displaystyle 3x-1\).
 

paulmdrdo

Active member
May 13, 2013
386
Re: integrals giving inverse trig.

yes i missed that one.

my new answer is $\displaystyle\frac{3}{4}\ln\left(x+\frac{\sqrt{5}+1}{2}\right)-\frac{5}{2\sqrt{5}}\arctan\frac{2x+1}{\sqrt{5}}+C$

is this correct?
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: integrals giving inverse trig.

You have the term \(\displaystyle \ln(a^2+u^2)\)

what do you get by letting

\(\displaystyle a^2=\frac{5}{4},u=x+\frac{1}{2}\)
 

paulmdrdo

Active member
May 13, 2013
386
Re: integrals giving inverse trig.

You have the term \(\displaystyle \ln(a^2+u^2)\)

what do you get by letting

\(\displaystyle a^2=\frac{5}{4},u=x+\frac{1}{2}\)
oh hey, i forgot to square them

$\displaystyle\frac{3}{4}\ln\left(x^2+x+\frac{3}{2}\right)-\frac{5}{2\sqrt{5}}\arctan\frac{2x+1}{\sqrt{5}}+C$ will this be the correct answer?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: integrals giving inverse trig.

oh hey, i forgot to square them

$\displaystyle\frac{3}{4}\ln\left(x^2+x+\frac{3}{2}\right)-\frac{5}{2\sqrt{5}}\arctan\frac{2x+1}{\sqrt{5}}+C$ will this be the correct answer?
Yes , sure .
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
again, i need some help here guys.


$\displaystyle\int\frac{3x-1}{2x^2+2x+3}dx$

=$\displaystyle\int\frac{3x-1}{2\left[\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\right]}dx$

$\displaystyle a=\frac{\sqrt{5}}{2}$; $\displaystyle u=x+\frac{1}{2}$; $\displaystyle du=dx$; $\displaystyle x=u-\frac{1}{2}$

=$\displaystyle\frac{1}{2}\int3\frac{\left(u-\frac{1}{2}\right)}{a^2+u^2}du$;

=$\displaystyle\frac{1}{2}\int\frac{\left(3u-\frac{3}{2}\right)}{a^2+u^2}du$

=$\displaystyle\frac{1}{2}\left(\int\frac{3u}{a^2+u^2}du-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

$\displaystyle v=a^2+u^2$; $\displaystyle dv=2udu$; $\displaystyle du=\frac{1}{2u}dv$

=$\displaystyle\frac{1}{2}\left(3\int\frac{udv}{v\,2u}dv-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

=$\displaystyle\frac{1}{2}\left(\frac{3}{2}\int \frac{dv}{v}-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

=$\displaystyle\frac{1}{2}\left(\frac{3}{2}ln(a^2+u^2)-\frac{3}{2}\frac{2}{\sqrt{5}}\arctan\frac{2x+1}{ \sqrt{5}}\right)$

=$\displaystyle\frac{3}{4}ln\left(x+\frac{\sqrt{5}+1}{2}\right)-\frac{3}{2\sqrt{5}}\arctan\frac{2x+1}{ \sqrt{5}}+C$ ---> FINAL ANSWER.

would you be so generous to check if my solution is correct?

thanks!

the answer in my book is in this form,

$\displaystyle\frac{3}{4}ln(2x^2+2x+3)-\frac{\sqrt{5}}{2}\arctan\frac{2x+1}{\sqrt{5}}+C$

is my answer equivalent to this one?
A good strategy is ALWAYS to look for a simple substitution, or to see how you can manipulate your integrand to get a simple substitution...

[tex]\displaystyle \begin{align*} \int{ \frac{3x - 1}{2x^2 + 2x + 3} \, dx} &= 3\int{\frac{x - \frac{1}{3}}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \int{ \frac{4x - \frac{4}{3}}{2x^2 + 2x + 3} \, dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2 - \frac{10}{3}}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \left( \int{ \frac{4x + 2}{2x^2 + 2x + 3}\,dx} - \frac{10}{3} \int{ \frac{1}{2x^2 + 2x + 3} \, dx} \right) \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3}\,dx} - \frac{5}{2} \int{ \frac{1}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \,dx } - \frac{5}{4} \int{ \frac{1}{x^2 + x + \frac{3}{2} }\,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \,dx} - \frac{5}{4} \int{ \frac{1}{x^2 + x + \left( \frac{1}{2} \right) ^2 - \left( \frac{1}{2} \right) ^2 + \frac{6}{4} } \,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \, dx} - \frac{5}{4} \int{ \frac{1}{ \left( x + \frac{1}{2} \right) ^2 + \frac{5}{4} } \, dx} \end{align*}[/tex]

The first of these integrals can now be solved using the substitution [tex]\displaystyle \begin{align*} u = 2x^2 + 2x + 3 \implies du = \left( 4x + 2 \right) \, dx \end{align*}[/tex] and the second can be solved using the substitution [tex]\displaystyle \begin{align*} x + \frac{1}{2} = \frac{\sqrt{5}}{2} \tan{(\theta)} \implies dx = \frac{\sqrt{5}}{2} \sec^2{(\theta)}\,d\theta \end{align*}[/tex].
 

paulmdrdo

Active member
May 13, 2013
386
hey zaid, for educational purposes could you tell me why the answer in my book is in this form,

$\displaystyle\frac{3}{4}ln(2x^2+2x+3)-\frac{\sqrt{5}}{2}\arctan\frac{2x+1}{\sqrt{5}}+C$

is this equivalent to what i get in my answer?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Yes, they rationalised the denominator.

[tex]\displaystyle \begin{align*} \frac{5}{2\sqrt{5}} &= \frac{5}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} \\ &= \frac{5\sqrt{5}}{2 \cdot 5} \\ &= \frac{\sqrt{5}}{2} \end{align*}[/tex]
 

paulmdrdo

Active member
May 13, 2013
386
A good strategy is ALWAYS to look for a simple substitution, or to see how you can manipulate your integrand to get a simple substitution...

[tex]\displaystyle \begin{align*} \int{ \frac{3x - 1}{2x^2 + 2x + 3} \, dx} &= 3\int{\frac{x - \frac{1}{3}}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \int{ \frac{4x - \frac{4}{3}}{2x^2 + 2x + 3} \, dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2 - \frac{10}{3}}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \left( \int{ \frac{4x + 2}{2x^2 + 2x + 3}\,dx} - \frac{10}{3} \int{ \frac{1}{2x^2 + 2x + 3} \, dx} \right) \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3}\,dx} - \frac{5}{2} \int{ \frac{1}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \,dx } - \frac{5}{4} \int{ \frac{1}{x^2 + x + \frac{3}{2} }\,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \,dx} - \frac{5}{4} \int{ \frac{1}{x^2 + x + \left( \frac{1}{2} \right) ^2 - \left( \frac{1}{2} \right) ^2 + \frac{6}{4} } \,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \, dx} - \frac{5}{4} \int{ \frac{1}{ \left( x + \frac{1}{2} \right) ^2 + \frac{5}{4} } \, dx} \end{align*}[/tex]

The first of these integrals can now be solved using the substitution [tex]\displaystyle \begin{align*} u = 2x^2 + 2x + 3 \implies du = \left( 4x + 2 \right) \, dx \end{align*}[/tex] and the second can be solved using the substitution [tex]\displaystyle \begin{align*} x + \frac{1}{2} = \frac{\sqrt{5}}{2} \tan{(\theta)} \implies dx = \frac{\sqrt{5}}{2} \sec^2{(\theta)}\,d\theta \end{align*}[/tex].
hey prove it,

can you tell me how did you arrive at 2nd line of your solution?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403

paulmdrdo

Active member
May 13, 2013
386
Yes, they rationalised the denominator.

[tex]\displaystyle \begin{align*} \frac{5}{2\sqrt{5}} &= \frac{5}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} \\ &= \frac{5\sqrt{5}}{2 \cdot 5} \\ &= \frac{\sqrt{5}}{2} \end{align*}[/tex]
oh yes, but what about the $\ln(2x^2+2x+3)$?

- - - Updated - - -

Multiplying by 4?
hmm. what made you decide that you have to multiply it by 4? wouldn't that change the integrand?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
oh yes, but what about the $\ln(2x^2+2x+3)$?
You can write it as [tex]\displaystyle \begin{align*} \ln{ \left[ 2 \left( x^2 + x + \frac{3}{2} \right) \right] } = \ln{(2)} + \ln{ \left( x^2 + x + \frac{3}{2} \right) } \end{align*}[/tex], and since [tex]\displaystyle \begin{align*} \ln{(2)} \end{align*}[/tex] is a constant, it can combine with the integration constant.

hmm. what made you decide that you have to multiply it by 4? wouldn't that change the integrand?
I multiplied by 4 because I realised that if we make the simplest substitution [tex]\displaystyle \begin{align*} u = 2x^2 + 2x + 3 \end{align*}[/tex] its derivative [tex]\displaystyle \begin{align*} 4x + 2 \end{align*}[/tex] has to be a factor.

No, it didn't change the integrand, because I also divided by 4 outside.
 

paulmdrdo

Active member
May 13, 2013
386
No, it didn't change the integrand, because I also divided by 4 outside.
you mean like this,

$\displaystyle 3\int\frac{\left(x+\frac{1}{3}\right)}{2x^2+2x+3} \frac{4}{4}$ ?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
hey zaid, for educational purposes could you tell me why the answer in my book is in this form,

$\displaystyle\frac{3}{4}ln(2x^2+2x+3)-\frac{\sqrt{5}}{2}\arctan\frac{2x+1}{\sqrt{5}}+C$

is this equivalent to what i get in my answer?
This a common question and this is due to the constant of integration in that sense and since we are working on an indefinite integral let us take the following example :

\(\displaystyle \int \frac{1}{2x+1}\, dx \)

First approach :

\(\displaystyle \frac{1}{2}\int \frac{2}{2x+1}\, dx =\frac{1}{2}\ln(2x+1)+C\)

Second approach :

\(\displaystyle \frac{1}{2}\int \frac{1}{x+\frac{1}{2}}\, dx = \frac{1}{2}\ln\left(x+\frac{1}{2} \right) +C \)

Why the answers differ ? are my approaches correct ?