A very strange and difficult hyperbolic integral

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5












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This integral gave me serious problems, I tried to solve it by parts but it is madness! The calculations are too long and difficult, I do not think we should solve this.



$$int _ -frac 1 3 ^ frac 1 3 sqrt 36 x ^ 4 -40 x ^ 2 +4 cosh (3x+tanh ^ -1 (3x)-tanh ^ -1 (x)) $$



the answer must be $$frac 12+4 e ^ 2 9e $$
Can some expert help me?










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  • 1




    $begingroup$
    Mathematica confirms your answer.
    $endgroup$
    – David G. Stork
    Jan 10 at 22:23










  • $begingroup$
    I have not solved the integral, the final result is given in my text question
    $endgroup$
    – Leprep98
    Jan 10 at 22:27










  • $begingroup$
    @symchdmath I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
    $endgroup$
    – Leprep98
    Jan 11 at 13:45










  • $begingroup$
    Leprep98: Of course the integral with limits $pm 1/3$ will not equal the integral with $pm 2$. What value for the integral with the latter limits is "wrong"?
    $endgroup$
    – David G. Stork
    Jan 11 at 19:30











  • $begingroup$
    with maxima if I integrate my integral in [-2,2] i get 1172.899, if I integrate 2xsinh(3x)+(1-3x^2)cosh(3x) in [-2,2] i get -582.71. So it's not the same. Why?
    $endgroup$
    – Leprep98
    Jan 12 at 10:31















5












$begingroup$


This integral gave me serious problems, I tried to solve it by parts but it is madness! The calculations are too long and difficult, I do not think we should solve this.



$$int _ -frac 1 3 ^ frac 1 3 sqrt 36 x ^ 4 -40 x ^ 2 +4 cosh (3x+tanh ^ -1 (3x)-tanh ^ -1 (x)) $$



the answer must be $$frac 12+4 e ^ 2 9e $$
Can some expert help me?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Mathematica confirms your answer.
    $endgroup$
    – David G. Stork
    Jan 10 at 22:23










  • $begingroup$
    I have not solved the integral, the final result is given in my text question
    $endgroup$
    – Leprep98
    Jan 10 at 22:27










  • $begingroup$
    @symchdmath I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
    $endgroup$
    – Leprep98
    Jan 11 at 13:45










  • $begingroup$
    Leprep98: Of course the integral with limits $pm 1/3$ will not equal the integral with $pm 2$. What value for the integral with the latter limits is "wrong"?
    $endgroup$
    – David G. Stork
    Jan 11 at 19:30











  • $begingroup$
    with maxima if I integrate my integral in [-2,2] i get 1172.899, if I integrate 2xsinh(3x)+(1-3x^2)cosh(3x) in [-2,2] i get -582.71. So it's not the same. Why?
    $endgroup$
    – Leprep98
    Jan 12 at 10:31













5












5








5





$begingroup$


This integral gave me serious problems, I tried to solve it by parts but it is madness! The calculations are too long and difficult, I do not think we should solve this.



$$int _ -frac 1 3 ^ frac 1 3 sqrt 36 x ^ 4 -40 x ^ 2 +4 cosh (3x+tanh ^ -1 (3x)-tanh ^ -1 (x)) $$



the answer must be $$frac 12+4 e ^ 2 9e $$
Can some expert help me?










share|cite|improve this question











$endgroup$




This integral gave me serious problems, I tried to solve it by parts but it is madness! The calculations are too long and difficult, I do not think we should solve this.



$$int _ -frac 1 3 ^ frac 1 3 sqrt 36 x ^ 4 -40 x ^ 2 +4 cosh (3x+tanh ^ -1 (3x)-tanh ^ -1 (x)) $$



the answer must be $$frac 12+4 e ^ 2 9e $$
Can some expert help me?







calculus integration






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 23:01







Leprep98

















asked Jan 10 at 22:13









Leprep98Leprep98

578




578







  • 1




    $begingroup$
    Mathematica confirms your answer.
    $endgroup$
    – David G. Stork
    Jan 10 at 22:23










  • $begingroup$
    I have not solved the integral, the final result is given in my text question
    $endgroup$
    – Leprep98
    Jan 10 at 22:27










  • $begingroup$
    @symchdmath I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
    $endgroup$
    – Leprep98
    Jan 11 at 13:45










  • $begingroup$
    Leprep98: Of course the integral with limits $pm 1/3$ will not equal the integral with $pm 2$. What value for the integral with the latter limits is "wrong"?
    $endgroup$
    – David G. Stork
    Jan 11 at 19:30











  • $begingroup$
    with maxima if I integrate my integral in [-2,2] i get 1172.899, if I integrate 2xsinh(3x)+(1-3x^2)cosh(3x) in [-2,2] i get -582.71. So it's not the same. Why?
    $endgroup$
    – Leprep98
    Jan 12 at 10:31












  • 1




    $begingroup$
    Mathematica confirms your answer.
    $endgroup$
    – David G. Stork
    Jan 10 at 22:23










  • $begingroup$
    I have not solved the integral, the final result is given in my text question
    $endgroup$
    – Leprep98
    Jan 10 at 22:27










  • $begingroup$
    @symchdmath I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
    $endgroup$
    – Leprep98
    Jan 11 at 13:45










  • $begingroup$
    Leprep98: Of course the integral with limits $pm 1/3$ will not equal the integral with $pm 2$. What value for the integral with the latter limits is "wrong"?
    $endgroup$
    – David G. Stork
    Jan 11 at 19:30











  • $begingroup$
    with maxima if I integrate my integral in [-2,2] i get 1172.899, if I integrate 2xsinh(3x)+(1-3x^2)cosh(3x) in [-2,2] i get -582.71. So it's not the same. Why?
    $endgroup$
    – Leprep98
    Jan 12 at 10:31







1




1




$begingroup$
Mathematica confirms your answer.
$endgroup$
– David G. Stork
Jan 10 at 22:23




$begingroup$
Mathematica confirms your answer.
$endgroup$
– David G. Stork
Jan 10 at 22:23












$begingroup$
I have not solved the integral, the final result is given in my text question
$endgroup$
– Leprep98
Jan 10 at 22:27




$begingroup$
I have not solved the integral, the final result is given in my text question
$endgroup$
– Leprep98
Jan 10 at 22:27












$begingroup$
@symchdmath I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
$endgroup$
– Leprep98
Jan 11 at 13:45




$begingroup$
@symchdmath I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
$endgroup$
– Leprep98
Jan 11 at 13:45












$begingroup$
Leprep98: Of course the integral with limits $pm 1/3$ will not equal the integral with $pm 2$. What value for the integral with the latter limits is "wrong"?
$endgroup$
– David G. Stork
Jan 11 at 19:30





$begingroup$
Leprep98: Of course the integral with limits $pm 1/3$ will not equal the integral with $pm 2$. What value for the integral with the latter limits is "wrong"?
$endgroup$
– David G. Stork
Jan 11 at 19:30













$begingroup$
with maxima if I integrate my integral in [-2,2] i get 1172.899, if I integrate 2xsinh(3x)+(1-3x^2)cosh(3x) in [-2,2] i get -582.71. So it's not the same. Why?
$endgroup$
– Leprep98
Jan 12 at 10:31




$begingroup$
with maxima if I integrate my integral in [-2,2] i get 1172.899, if I integrate 2xsinh(3x)+(1-3x^2)cosh(3x) in [-2,2] i get -582.71. So it's not the same. Why?
$endgroup$
– Leprep98
Jan 12 at 10:31










2 Answers
2






active

oldest

votes


















15












$begingroup$

This integral is more complicated than it looks and only requires comfort with hyperbolic trigonometric definitions. My initial instinct is to look at the expression inside the $cosh$ to see if I can simplify it. In fact we have by the definition of the hyperbolic trigonometric functions,



$$tanh^-1(x) = frac12 ln left(frac1+x1-xright) $$



Now using log laws we have that,



$$tanh^-1(3x) - tanh^-1(x) = frac12 ln left(frac(1+3x)(1-x)(1-3x)(1+x) right), (*)$$



We note at this point that,



$$sqrt36x^4 - 40x^2 + 4 = 2sqrt(1-9x^2)(1-x^2) = 2sqrt(1-3x)(1+3x)(1-x)(1+x)$$



There is a lot in common between the above two equations which motivates us to press forward, we now use the following identity which we can prove just with the definitions of the hyperbolic trig functions,



$$cosh(x+y) = cosh(x) cosh(y) + sinh(x) sinh(y)$$



We use to simplify the $cosh$ in the integral, we find that,



$$cosh(3x + (tanh^-1(3x) - tanh^-1(x))) = cosh(3x) cosh(tanh^-1(3x) - tanh^-1(x)) + sinh(3x) sinh(tanh^-1(3x) - tanh^-1(x)) $$



Using the following definitions of the hyperbolic functions,



$$cosh(x) = frace^x + e^-x2 $$



$$sinh(x) = frace^x - e^-x2 $$



We find that using $(*)$, leaving the details to you,



$$cosh(tanh^-1(3x) - tanh^-1(x)) = coshleft(frac12ln left(frac(1+3x)(1-x)(1-3x)(1+x) right)right) = frac1-3x^2sqrt(1-9x^2)(1-x^2)$$



Similarly, we have



$$sinh(tanh^-1(3x) - tanh^-1(x)) = frac2xsqrt(1-9x^2)(1-x^2) $$



Putting this all together,



$$I = int_-1/3^1/3 sqrt36x^4 - 40x^2 + 4 cosh(3x + tanh^-1(3x) - tanh^-1(x)) mathrmdx $$



$$I = int_-1/3^1/3 2sqrt(1-9x^2)(1-x^2) left(cosh(3x) frac1-3x^2sqrt(1-9x^2)(1-x^2) + sinh(3x)frac2xsqrt(1-9x^2)(1-x^2) right) mathrmdx$$



Finally the integral simplifies to,



$$I = 2 int_-1/3^1/3 (1-3x^2)cosh(3x) mathrmdx + 2 int_-1/3^1/3 2x sinh(3x) mathrmdx $$



Which is a lot easier to compute with IBP and I will leave that to you to complete, this does indeed give the correct answer.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    note: 2xcosh (3x) is wrong, you should write 2xsinh (3x)
    $endgroup$
    – Leprep98
    Jan 11 at 12:56










  • $begingroup$
    I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
    $endgroup$
    – Leprep98
    Jan 11 at 13:46


















1












$begingroup$

Use the addition formula and simplify the $cosh(tanh^-1(cdot))$, $sinh(tanh^-1(cdot))$, your integrand becomes $2cosh(3x)(1-3x^2)+4xsinh(3x)$.



Note that $2cosh(3x)(1-3x^2)-4xsinh(3x)$ is the derivative of the function $frac23sinh(3x)(1-3x^2)$ so its integral is $frac49e(e^2-1)$.



So it remains to integrate $8xsinh(3x)$. By parts, this is $frac169cosh(1)-frac83intcosh(3x)=frac89e(e^2+1)-frac89e(e^2-1)$.



Summing yields finally $frac4e^2+129e$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    15












    $begingroup$

    This integral is more complicated than it looks and only requires comfort with hyperbolic trigonometric definitions. My initial instinct is to look at the expression inside the $cosh$ to see if I can simplify it. In fact we have by the definition of the hyperbolic trigonometric functions,



    $$tanh^-1(x) = frac12 ln left(frac1+x1-xright) $$



    Now using log laws we have that,



    $$tanh^-1(3x) - tanh^-1(x) = frac12 ln left(frac(1+3x)(1-x)(1-3x)(1+x) right), (*)$$



    We note at this point that,



    $$sqrt36x^4 - 40x^2 + 4 = 2sqrt(1-9x^2)(1-x^2) = 2sqrt(1-3x)(1+3x)(1-x)(1+x)$$



    There is a lot in common between the above two equations which motivates us to press forward, we now use the following identity which we can prove just with the definitions of the hyperbolic trig functions,



    $$cosh(x+y) = cosh(x) cosh(y) + sinh(x) sinh(y)$$



    We use to simplify the $cosh$ in the integral, we find that,



    $$cosh(3x + (tanh^-1(3x) - tanh^-1(x))) = cosh(3x) cosh(tanh^-1(3x) - tanh^-1(x)) + sinh(3x) sinh(tanh^-1(3x) - tanh^-1(x)) $$



    Using the following definitions of the hyperbolic functions,



    $$cosh(x) = frace^x + e^-x2 $$



    $$sinh(x) = frace^x - e^-x2 $$



    We find that using $(*)$, leaving the details to you,



    $$cosh(tanh^-1(3x) - tanh^-1(x)) = coshleft(frac12ln left(frac(1+3x)(1-x)(1-3x)(1+x) right)right) = frac1-3x^2sqrt(1-9x^2)(1-x^2)$$



    Similarly, we have



    $$sinh(tanh^-1(3x) - tanh^-1(x)) = frac2xsqrt(1-9x^2)(1-x^2) $$



    Putting this all together,



    $$I = int_-1/3^1/3 sqrt36x^4 - 40x^2 + 4 cosh(3x + tanh^-1(3x) - tanh^-1(x)) mathrmdx $$



    $$I = int_-1/3^1/3 2sqrt(1-9x^2)(1-x^2) left(cosh(3x) frac1-3x^2sqrt(1-9x^2)(1-x^2) + sinh(3x)frac2xsqrt(1-9x^2)(1-x^2) right) mathrmdx$$



    Finally the integral simplifies to,



    $$I = 2 int_-1/3^1/3 (1-3x^2)cosh(3x) mathrmdx + 2 int_-1/3^1/3 2x sinh(3x) mathrmdx $$



    Which is a lot easier to compute with IBP and I will leave that to you to complete, this does indeed give the correct answer.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      note: 2xcosh (3x) is wrong, you should write 2xsinh (3x)
      $endgroup$
      – Leprep98
      Jan 11 at 12:56










    • $begingroup$
      I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
      $endgroup$
      – Leprep98
      Jan 11 at 13:46















    15












    $begingroup$

    This integral is more complicated than it looks and only requires comfort with hyperbolic trigonometric definitions. My initial instinct is to look at the expression inside the $cosh$ to see if I can simplify it. In fact we have by the definition of the hyperbolic trigonometric functions,



    $$tanh^-1(x) = frac12 ln left(frac1+x1-xright) $$



    Now using log laws we have that,



    $$tanh^-1(3x) - tanh^-1(x) = frac12 ln left(frac(1+3x)(1-x)(1-3x)(1+x) right), (*)$$



    We note at this point that,



    $$sqrt36x^4 - 40x^2 + 4 = 2sqrt(1-9x^2)(1-x^2) = 2sqrt(1-3x)(1+3x)(1-x)(1+x)$$



    There is a lot in common between the above two equations which motivates us to press forward, we now use the following identity which we can prove just with the definitions of the hyperbolic trig functions,



    $$cosh(x+y) = cosh(x) cosh(y) + sinh(x) sinh(y)$$



    We use to simplify the $cosh$ in the integral, we find that,



    $$cosh(3x + (tanh^-1(3x) - tanh^-1(x))) = cosh(3x) cosh(tanh^-1(3x) - tanh^-1(x)) + sinh(3x) sinh(tanh^-1(3x) - tanh^-1(x)) $$



    Using the following definitions of the hyperbolic functions,



    $$cosh(x) = frace^x + e^-x2 $$



    $$sinh(x) = frace^x - e^-x2 $$



    We find that using $(*)$, leaving the details to you,



    $$cosh(tanh^-1(3x) - tanh^-1(x)) = coshleft(frac12ln left(frac(1+3x)(1-x)(1-3x)(1+x) right)right) = frac1-3x^2sqrt(1-9x^2)(1-x^2)$$



    Similarly, we have



    $$sinh(tanh^-1(3x) - tanh^-1(x)) = frac2xsqrt(1-9x^2)(1-x^2) $$



    Putting this all together,



    $$I = int_-1/3^1/3 sqrt36x^4 - 40x^2 + 4 cosh(3x + tanh^-1(3x) - tanh^-1(x)) mathrmdx $$



    $$I = int_-1/3^1/3 2sqrt(1-9x^2)(1-x^2) left(cosh(3x) frac1-3x^2sqrt(1-9x^2)(1-x^2) + sinh(3x)frac2xsqrt(1-9x^2)(1-x^2) right) mathrmdx$$



    Finally the integral simplifies to,



    $$I = 2 int_-1/3^1/3 (1-3x^2)cosh(3x) mathrmdx + 2 int_-1/3^1/3 2x sinh(3x) mathrmdx $$



    Which is a lot easier to compute with IBP and I will leave that to you to complete, this does indeed give the correct answer.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      note: 2xcosh (3x) is wrong, you should write 2xsinh (3x)
      $endgroup$
      – Leprep98
      Jan 11 at 12:56










    • $begingroup$
      I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
      $endgroup$
      – Leprep98
      Jan 11 at 13:46













    15












    15








    15





    $begingroup$

    This integral is more complicated than it looks and only requires comfort with hyperbolic trigonometric definitions. My initial instinct is to look at the expression inside the $cosh$ to see if I can simplify it. In fact we have by the definition of the hyperbolic trigonometric functions,



    $$tanh^-1(x) = frac12 ln left(frac1+x1-xright) $$



    Now using log laws we have that,



    $$tanh^-1(3x) - tanh^-1(x) = frac12 ln left(frac(1+3x)(1-x)(1-3x)(1+x) right), (*)$$



    We note at this point that,



    $$sqrt36x^4 - 40x^2 + 4 = 2sqrt(1-9x^2)(1-x^2) = 2sqrt(1-3x)(1+3x)(1-x)(1+x)$$



    There is a lot in common between the above two equations which motivates us to press forward, we now use the following identity which we can prove just with the definitions of the hyperbolic trig functions,



    $$cosh(x+y) = cosh(x) cosh(y) + sinh(x) sinh(y)$$



    We use to simplify the $cosh$ in the integral, we find that,



    $$cosh(3x + (tanh^-1(3x) - tanh^-1(x))) = cosh(3x) cosh(tanh^-1(3x) - tanh^-1(x)) + sinh(3x) sinh(tanh^-1(3x) - tanh^-1(x)) $$



    Using the following definitions of the hyperbolic functions,



    $$cosh(x) = frace^x + e^-x2 $$



    $$sinh(x) = frace^x - e^-x2 $$



    We find that using $(*)$, leaving the details to you,



    $$cosh(tanh^-1(3x) - tanh^-1(x)) = coshleft(frac12ln left(frac(1+3x)(1-x)(1-3x)(1+x) right)right) = frac1-3x^2sqrt(1-9x^2)(1-x^2)$$



    Similarly, we have



    $$sinh(tanh^-1(3x) - tanh^-1(x)) = frac2xsqrt(1-9x^2)(1-x^2) $$



    Putting this all together,



    $$I = int_-1/3^1/3 sqrt36x^4 - 40x^2 + 4 cosh(3x + tanh^-1(3x) - tanh^-1(x)) mathrmdx $$



    $$I = int_-1/3^1/3 2sqrt(1-9x^2)(1-x^2) left(cosh(3x) frac1-3x^2sqrt(1-9x^2)(1-x^2) + sinh(3x)frac2xsqrt(1-9x^2)(1-x^2) right) mathrmdx$$



    Finally the integral simplifies to,



    $$I = 2 int_-1/3^1/3 (1-3x^2)cosh(3x) mathrmdx + 2 int_-1/3^1/3 2x sinh(3x) mathrmdx $$



    Which is a lot easier to compute with IBP and I will leave that to you to complete, this does indeed give the correct answer.






    share|cite|improve this answer











    $endgroup$



    This integral is more complicated than it looks and only requires comfort with hyperbolic trigonometric definitions. My initial instinct is to look at the expression inside the $cosh$ to see if I can simplify it. In fact we have by the definition of the hyperbolic trigonometric functions,



    $$tanh^-1(x) = frac12 ln left(frac1+x1-xright) $$



    Now using log laws we have that,



    $$tanh^-1(3x) - tanh^-1(x) = frac12 ln left(frac(1+3x)(1-x)(1-3x)(1+x) right), (*)$$



    We note at this point that,



    $$sqrt36x^4 - 40x^2 + 4 = 2sqrt(1-9x^2)(1-x^2) = 2sqrt(1-3x)(1+3x)(1-x)(1+x)$$



    There is a lot in common between the above two equations which motivates us to press forward, we now use the following identity which we can prove just with the definitions of the hyperbolic trig functions,



    $$cosh(x+y) = cosh(x) cosh(y) + sinh(x) sinh(y)$$



    We use to simplify the $cosh$ in the integral, we find that,



    $$cosh(3x + (tanh^-1(3x) - tanh^-1(x))) = cosh(3x) cosh(tanh^-1(3x) - tanh^-1(x)) + sinh(3x) sinh(tanh^-1(3x) - tanh^-1(x)) $$



    Using the following definitions of the hyperbolic functions,



    $$cosh(x) = frace^x + e^-x2 $$



    $$sinh(x) = frace^x - e^-x2 $$



    We find that using $(*)$, leaving the details to you,



    $$cosh(tanh^-1(3x) - tanh^-1(x)) = coshleft(frac12ln left(frac(1+3x)(1-x)(1-3x)(1+x) right)right) = frac1-3x^2sqrt(1-9x^2)(1-x^2)$$



    Similarly, we have



    $$sinh(tanh^-1(3x) - tanh^-1(x)) = frac2xsqrt(1-9x^2)(1-x^2) $$



    Putting this all together,



    $$I = int_-1/3^1/3 sqrt36x^4 - 40x^2 + 4 cosh(3x + tanh^-1(3x) - tanh^-1(x)) mathrmdx $$



    $$I = int_-1/3^1/3 2sqrt(1-9x^2)(1-x^2) left(cosh(3x) frac1-3x^2sqrt(1-9x^2)(1-x^2) + sinh(3x)frac2xsqrt(1-9x^2)(1-x^2) right) mathrmdx$$



    Finally the integral simplifies to,



    $$I = 2 int_-1/3^1/3 (1-3x^2)cosh(3x) mathrmdx + 2 int_-1/3^1/3 2x sinh(3x) mathrmdx $$



    Which is a lot easier to compute with IBP and I will leave that to you to complete, this does indeed give the correct answer.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 12 at 10:01









    Leprep98

    578




    578










    answered Jan 10 at 23:42









    symchdmathsymchdmath

    39816




    39816











    • $begingroup$
      note: 2xcosh (3x) is wrong, you should write 2xsinh (3x)
      $endgroup$
      – Leprep98
      Jan 11 at 12:56










    • $begingroup$
      I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
      $endgroup$
      – Leprep98
      Jan 11 at 13:46
















    • $begingroup$
      note: 2xcosh (3x) is wrong, you should write 2xsinh (3x)
      $endgroup$
      – Leprep98
      Jan 11 at 12:56










    • $begingroup$
      I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
      $endgroup$
      – Leprep98
      Jan 11 at 13:46















    $begingroup$
    note: 2xcosh (3x) is wrong, you should write 2xsinh (3x)
    $endgroup$
    – Leprep98
    Jan 11 at 12:56




    $begingroup$
    note: 2xcosh (3x) is wrong, you should write 2xsinh (3x)
    $endgroup$
    – Leprep98
    Jan 11 at 12:56












    $begingroup$
    I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
    $endgroup$
    – Leprep98
    Jan 11 at 13:46




    $begingroup$
    I check your result with Maxima and the solution it's correct, but whit another interval of integration for example [-2,2] the result it's not correct. Can you explain me why?
    $endgroup$
    – Leprep98
    Jan 11 at 13:46











    1












    $begingroup$

    Use the addition formula and simplify the $cosh(tanh^-1(cdot))$, $sinh(tanh^-1(cdot))$, your integrand becomes $2cosh(3x)(1-3x^2)+4xsinh(3x)$.



    Note that $2cosh(3x)(1-3x^2)-4xsinh(3x)$ is the derivative of the function $frac23sinh(3x)(1-3x^2)$ so its integral is $frac49e(e^2-1)$.



    So it remains to integrate $8xsinh(3x)$. By parts, this is $frac169cosh(1)-frac83intcosh(3x)=frac89e(e^2+1)-frac89e(e^2-1)$.



    Summing yields finally $frac4e^2+129e$.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      Use the addition formula and simplify the $cosh(tanh^-1(cdot))$, $sinh(tanh^-1(cdot))$, your integrand becomes $2cosh(3x)(1-3x^2)+4xsinh(3x)$.



      Note that $2cosh(3x)(1-3x^2)-4xsinh(3x)$ is the derivative of the function $frac23sinh(3x)(1-3x^2)$ so its integral is $frac49e(e^2-1)$.



      So it remains to integrate $8xsinh(3x)$. By parts, this is $frac169cosh(1)-frac83intcosh(3x)=frac89e(e^2+1)-frac89e(e^2-1)$.



      Summing yields finally $frac4e^2+129e$.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        Use the addition formula and simplify the $cosh(tanh^-1(cdot))$, $sinh(tanh^-1(cdot))$, your integrand becomes $2cosh(3x)(1-3x^2)+4xsinh(3x)$.



        Note that $2cosh(3x)(1-3x^2)-4xsinh(3x)$ is the derivative of the function $frac23sinh(3x)(1-3x^2)$ so its integral is $frac49e(e^2-1)$.



        So it remains to integrate $8xsinh(3x)$. By parts, this is $frac169cosh(1)-frac83intcosh(3x)=frac89e(e^2+1)-frac89e(e^2-1)$.



        Summing yields finally $frac4e^2+129e$.






        share|cite|improve this answer









        $endgroup$



        Use the addition formula and simplify the $cosh(tanh^-1(cdot))$, $sinh(tanh^-1(cdot))$, your integrand becomes $2cosh(3x)(1-3x^2)+4xsinh(3x)$.



        Note that $2cosh(3x)(1-3x^2)-4xsinh(3x)$ is the derivative of the function $frac23sinh(3x)(1-3x^2)$ so its integral is $frac49e(e^2-1)$.



        So it remains to integrate $8xsinh(3x)$. By parts, this is $frac169cosh(1)-frac83intcosh(3x)=frac89e(e^2+1)-frac89e(e^2-1)$.



        Summing yields finally $frac4e^2+129e$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 23:56









        MindlackMindlack

        3,27717




        3,27717



























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