The integral \(\int(1+x^{2})^{\frac{3}{2}}dx\) can be effectively solved using hyperbolic substitution by letting \(x = \sinh(u)\), which leads to \(dx = \cosh(u) du\). This transforms the integral into \(\int\cosh^{4}(u) du\). The discussion highlights the importance of using the identity \(\cosh^{2}(t) = \frac{1 + \cosh(2t)}{2}\) for simplifying the evaluation of the integral. Additionally, a correction in LaTeX formatting was noted, emphasizing the need for proper syntax in mathematical expressions.
PREREQUISITESStudents studying calculus, particularly those focusing on integration techniques, as well as educators teaching hyperbolic functions and their applications in solving integrals.
<br /> <br /> Oops, that was meant to be nice neat LaTeX...<br /> <br /> Well, I tried doing that but ended up with int(cosh^4(u))du... not sure if there is a nice little trick for solving that or if I've ended up with a harder problem than I started with. Also, why didn't that tex code just work?Malby said:\begin{eqnarray*}<br /> \int\left(1+x^{2}\right)^{\frac{3}{2}}dx & = & \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du\\<br /> & = & \int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du\\<br /> & = & \int\sqrt{\cosh^{6}\left(u\right)}\cosh\left(u\right)\, du\\<br /> & = & \int\cosh^{4}\left(u\right)\, du<br /> \end{eqnarray*}[\tex]
Changed '\' to '/' in the /tex tag so we can read your LaTeX.Malby said:\int\left(1+x^{2}\right)^{\frac{3}{2}}dx = \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du
=\int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du
=\int\sqrt{\cosh^{6}\left(u\right)} \cosh\left(u\right)\, du
=\int\cosh^{4}\left(u\right)\, du<br />
By The Way: u = \sinh^{-1}(x) = \ln \left( x+\sqrt{x^{2}+1} \right)SammyS said:sinh2(u) + 1 = cosh2(u)
So, I would try letting x = sinh(u) → dx = cosh(u) du
SammyS said:By The Way: u = \sinh^{-1}(x) = \ln \left( x+\sqrt{x^{2}+1} \right)