Using calculus and trigonometry identities to find value A,B and C.

Dhamnekar Winod

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There exist positive integers A, B, C such that

\[\displaystyle\int_1^\sqrt{2} \frac1x\sqrt{\frac{4*x^2+1}{x^2+1}}dx=\frac1A\pi+\ln{B}-\tan^{-1}C.\]

What is the value of A+B+C?

Solution:-

I don't have any clue to solve this question. I only know the value of this integral is \(0.575038390411.\)

If any member knows the answer to this question, may reply with correct answer.
 

Dhamnekar Winod

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There exist positive integers A, B, C such that

\[\displaystyle\int_1^\sqrt{2} \frac1x\sqrt{\frac{4*x^2+1}{x^2+1}}dx=\frac1A\pi+\ln{B}-\tan^{-1}C.\]

What is the value of A+B+C?

Solution:-

I don't have any clue to solve this question. I only know the value of this integral is \(0.575038390411.\)

If any member knows the answer to this question, may reply with correct answer.
Please read L.H.S. of the equation as \[\displaystyle\int_1^\sqrt{2} \frac1x\sqrt{\frac{4*x^2+1}{x^2-1}}dx=\frac1A\pi+\ln{B}-\tan^{-1}C.\]
 

Dhamnekar Winod

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Please read L.H.S. of the equation as \[\displaystyle\int_1^\sqrt{2} \frac1x\sqrt{\frac{4*x^2+1}{x^2-1}}dx=\frac1A\pi+\ln{B}-\tan^{-1}C.\]
The value of this integral is stated wrong in my original post. Its correct value is \(1.93117201743\)
 

Jomo

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The exact value is not really going to help unless you cheat, Just let pi/A= 1.93117201743 so A = pi/1.93117201743, B=1 since ln(1) = 0 and C=pi. But this is nonsense especially since I doubt that the exact answer is 1.93117201743

You need to make some substitutions to get pi, ln(k) and tan-1m.

What have you tried? Where are you stuck? You need to show us something so we know how to help you. Or do you just want someone here to just solve this for you?
 
Last edited:

lev888

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The exact value is not really going to help unless you cheat, Just let pi/A= 1.93117201743 so A = pi/1.93117201743, B=1 since ln(1) = 0 and C=0. But this is nonsense especially since I doubt that the exact answer is 1.93117201743

You need to make some substitutions to get pi, ln(k) and tan-1m.

What have you tried? Where are you stuck? You need to show us something so we know how to help you. Or do you just want someone here to just solve this for you?
"There exist positive integers A, B, C..."
 

Jomo

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Dhamnekar Winod

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The answer to this question is A=2, B=5,and C=3, So, A+B+C=10.

The value of this integral is 1.931188466830743=
\(\displaystyle \frac{\pi}{2}+\ln{(5)}-\arctan{(3)}\)
 

Subhotosh Khan

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There exist positive integers A, B, C such that

\[\displaystyle\int_1^\sqrt{2} \frac1x\sqrt{\frac{4*x^2+1}{x^2+1}}dx=\frac1A\pi+\ln{B}-\tan^{-1}C.\]

What is the value of A+B+C?

Solution:-

I don't have any clue to solve this question. I only know the value of this integral is \(0.575038390411.\)

If any member knows the answer to this question, may reply with correct answer.
I like this problem - this where WA will not be able provide answer (in one step).

First solve the indefinite integral of the LHS.

\(\displaystyle \displaystyle\int \frac1x\sqrt{\frac{4*x^2+1}{x^2+1}}dx =? \)
 

Dhamnekar Winod

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I like this problem - this where WA will not be able provide answer (in one step).

First solve the indefinite integral of the LHS.

\(\displaystyle \displaystyle\int \frac1x\sqrt{\frac{4*x^2+1}{x^2+1}}dx =? \)
Let x=csc(t),so,

\(\displaystyle\int_1^{\sqrt{2}}\frac1x\sqrt{\frac{4*x^2+1}{x^2-1}}dx=\displaystyle\int^{\frac{\pi}{2}}_{\frac14*\pi} \sqrt{4*\csc^2{(t)}+1}dt\)


Now, solving R.H.S. of this equation, we get 1.931188466830743.
 
Last edited:

Subhotosh Khan

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Let x=csc(t),so,

\(\displaystyle\int_1^{\sqrt{2}}\frac1x\sqrt{\frac{4*x^2+1}{x^2-1}}dx=\displaystyle\int^{\frac{\pi}{2}}_{\frac14*\pi} \sqrt{4*\csc^2{(t)}+1}dt\)
Now, solving R.H.S. of this equation, we get 1.931188466830743.
But to slove the problem of OP, you need to:

First solve the indefinite integral of the LHS.
 

Jomo

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Let x=csc(t),so,

\(\displaystyle\int_1^{\sqrt{2}}\frac1x\sqrt{\frac{4*x^2+1}{x^2-1}}dx=\displaystyle\int^{\frac{\pi}{2}}_{\frac14*\pi} \sqrt{4*\csc^2{(t)}+1}dt\)


Now, solving R.H.S. of this equation, we get 1.931188466830743.
Why bother making the substitution if you already knew the answer was 1.931188466830743. Did you expect to get something different?
You need to continue and solve one of these indefinite integral.
 

Dhamnekar Winod

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Due to overwhelming response from the readers, viewers, members, visitors, guests, math experts of this www.freemathhelp.com/forum, I am posting the answer to this question for review of mathematical audience of this forum.

From the first step written in my previous post, I proceed further,


Put \(\csc{(x)}=\frac{\sec{(x)}}{\tan{(x)}}\) and \(\tan^2{(x)}=\sec^2{(x)}-1\)

\(\displaystyle\int\sqrt{4*\csc^2{(x)}+1}dx=\displaystyle\int\frac{\sqrt{5*\sec^2{(x)}-1}*\sec{(x)}\tan{(x)}}{\sec{(x)}*(\sec^2{(x)}-1)}dx ……... {(1)}\)



Substitute \(u=\sec{(x)}\), the L.H.S. of (1)=\(\displaystyle\int\frac{\sqrt{5*u^2-1}}{u*(u^2-1)}du\)

Substitute\(v=\sqrt{5*u^2-1}\), the L.H.S of (1) =\(5*\displaystyle\int\frac{v^2}{(v^2-4)*(v+1)}dv\)

So, L.H.S. of (1)=\(-\ln{(v+2)}+\arctan{(v)}+\ln{(v-2)}\)


Undoing the substitution \(v =\sqrt{5*u^2-1}\) and \(u=\sec{(x)}\)

The L.H.S.of (1)=\(-\ln{(\sqrt{5*\sec^2{(x)}-1}+2)}+\arctan{(\sqrt{(5*\sec^2{(x)}-1)})} + \ln{(\sqrt{5*\sec^2{(x)}-1}-2)}\)

Hence, the definite integral of the L.H.S.of (1) from \(\frac{\pi}{4}\) to \(\frac{\pi}{2} = \frac{\pi}{2} +\ln{(5)}-\arctan{(3)}\)

Hence A+B+C=10, where A=2, B=5 and C=3. (QED)
 

Jomo

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Your v-sub was done correctly. Also you left out a lot of work right after that.
 

Cubist

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Your v-sub was done correctly. Also you left out a lot of work right after that.
I've enjoyed trying to verify OP's work. He left out enough steps to make this excellent revision practice for me. I made a few mistakes en-route (thanks to the moderator for deleting some of my erroneous posts!)

I'm pretty sure that he made an error at the v-sub stage (probably a transcription error)...
\(\displaystyle 5\int \frac{v^2}{\left(v^2-4\right)\left(v^{\color{red}{2}}+1\right)}\,dv \)
...Jomo would you be willing to please re-check the v-sub stage with this in mind?

Then I used partial fractions to obtain something in the form...
\(\displaystyle \frac{A}{v+2}+\frac{B}{v^2+1}+\frac{C}{v-2}\)

which then integrates nicely to OP's next line.
 

Jomo

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Your v-sub was done correctly. Also you left out a lot of work right after that.
Edit: I meant to say that your v-sub was not done correctly.
 

Jomo

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I've enjoyed trying to verify OP's work. He left out enough steps to make this excellent revision practice for me. I made a few mistakes en-route (thanks to the moderator for deleting some of my erroneous posts!)

I'm pretty sure that he made an error at the v-sub stage (probably a transcription error)...
\(\displaystyle 5\int \frac{v^2}{\left(v^2-4\right)\left(v^{\color{red}{2}}+1\right)}\,dv \)
...Jomo would you be willing to please re-check the v-sub stage with this in mind?

Then I used partial fractions to obtain something in the form...
\(\displaystyle \frac{A}{v+2}+\frac{B}{v^2+1}+\frac{C}{v-2}\)

which then integrates nicely to OP's next line.
yeah yeah, I noticed that and meant to say that the v-sub was NOT correct.
 
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