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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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    $\begingroup$ 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
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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,

$$\tanh^{-1}(x) = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) $$

Now using log laws we have that,

$$\tanh^{-1}(3x) - \tanh^{-1}(x) = \frac{1}{2} \ln \left(\frac{(1+3x)(1-x)}{(1-3x)(1+x)} \right), \ \ (*)$$

We note at this point that,

$$\sqrt{36x^4 - 40x^2 + 4} = 2\sqrt{(1-9x^2)(1-x^2)} = 2\sqrt{(1-3x)(1+3x)(1-x)(1+x)}$$

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,

$$\cosh(x) = \frac{e^x + e^{-x}}{2} $$

$$\sinh(x) = \frac{e^x - e^{-x}}{2} $$

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

$$\cosh(\tanh^{-1}(3x) - \tanh^{-1}(x)) = \cosh\left(\frac{1}{2}\ln \left(\frac{(1+3x)(1-x)}{(1-3x)(1+x)} \right)\right) = \frac{1-3x^2}{\sqrt{(1-9x^2)(1-x^2)}}$$

Similarly, we have

$$\sinh(\tanh^{-1}(3x) - \tanh^{-1}(x)) = \frac{2x}{\sqrt{(1-9x^2)(1-x^2)}} $$

Putting this all together,

$$I = \int_{-1/3}^{1/3} \sqrt{36x^4 - 40x^2 + 4} \cosh(3x + \tanh^{-1}(3x) - \tanh^{-1}(x)) \ \mathrm{d}x $$

$$I = \int_{-1/3}^{1/3} 2\sqrt{(1-9x^2)(1-x^2)} \left(\cosh(3x) \frac{1-3x^2}{\sqrt{(1-9x^2)(1-x^2)}} + \sinh(3x)\frac{2x}{\sqrt{(1-9x^2)(1-x^2)}} \right) \ \mathrm{d}x$$

Finally the integral simplifies to,

$$I = 2 \int_{-1/3}^{1/3} (1-3x^2)\cosh(3x) \ \mathrm{d}x + 2 \int_{-1/3}^{1/3} 2x \sinh(3x) \ \mathrm{d}x $$

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.

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

Note that $2\cosh(3x)(1-3x^2)-4x\sinh(3x)$ is the derivative of the function $\frac{2}{3}\sinh(3x)(1-3x^2)$ so its integral is $\frac{4}{9e}(e^2-1)$.

So it remains to integrate $8x\sinh(3x)$. By parts, this is $\frac{16}{9}\cosh(1)-\frac{8}{3}\int{\cosh(3x)}=\frac{8}{9e}(e^2+1)-\frac{8}{9e}(e^2-1)$.

Summing yields finally $\frac{4e^2+12}{9e}$.

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