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Using the substitution $x=\cosh (t)$ or otherwise, find $$\int\frac{x^3}{\sqrt{x^2-1}}dx$$ The correct answer is apparently $$\frac{1}{3}\sqrt{x^2-1}(x^2+2)$$ I seem to have gone very wrong somewhere; my answer is way off, can someone explain how to get this answer to me.

Thanks.

My working: $$\int\frac{\cosh^3t}{\sinh^2t}dt$$ $$u=\sinh t$$ $$\int\frac{1+u^2}{u^2}du$$ $$\frac{-1}{u}+u$$ $$\frac{-1}{\sinh t}+\sinh t$$ $$\frac{-1}{\sqrt{x^2-1}}+\sqrt{x^2-1}$$

^my working, I'm pretty sure this is very wrong though.

Edit: I've spotted my error. On the first line it should be $$\int \cosh^3t \, dt$$

not

$$\int\frac{\cosh^3t}{\sinh^2t}dt$$

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  • $\begingroup$ Can you include the work you did up to the point you got stuck? $\endgroup$ – graydad Feb 1 '15 at 21:49
  • $\begingroup$ how about a u= x^2-1 substitution $\endgroup$ – Asier Calbet Feb 1 '15 at 22:00
  • $\begingroup$ @Hasan, I believe Peter is right $cosht^2 - 1 = sinht^2$ is under a radical, so the square root would just become $sinht$. $\endgroup$ – PawnInGameOfLife Feb 1 '15 at 22:07
  • $\begingroup$ But the $sinht$ cancels out because of $dx=sinht\ dt$, so the integrand becomes $cosh^3x$. $\endgroup$ – Peter Feb 1 '15 at 22:08
  • $\begingroup$ $\cosh^3x=\cosh^2x\cosh x$. Use substitution. $\endgroup$ – velut luna Feb 1 '15 at 22:10
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Trig substitutions are not necessary. You can substitute $u = x^2 - 1$ and get

$$\int \frac{x^3}{\sqrt{x^2-1}}\,dx = \int \frac{(x^2 - 1 + 1)}{\sqrt{x^2-1}} x\,dx = \frac{1}{2}\int \frac{u+1}{\sqrt{u}} \,du = \frac{1}{2} \int (u^{1/2} + u^{-1/2} )\,du $$

The key word is "or otherwise"

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If you want to do this by using the substitution $x = \cosh(t)$ you also need $dx = \sinh(t)dt$, which means $$\int \frac{x^3}{\sqrt{x^2-1}}dx = \int \frac{\cosh^3(t)\sinh(t)dt}{\sqrt{\cosh^2(t)-1}}$$ and now use $\cosh^2(t) = 1+\sinh^2(t)$ in the numerator to get $$\int \frac{\cosh^3(t)\sinh(t)dt}{\sqrt{\cosh^2(t-11}} = \int \frac{(1+\sinh^2(t))\cosh(t)\sinh(t)dt}{\sqrt{\cosh^2(t)-1}} \\ =\int \frac{\cosh(t)\sinh(t)+\cosh(t)\sinh^3(t)}{\sqrt{\cosh^2(t)-1}}dt \\ = \int \frac{\cosh(t)\sinh(t)}{\sqrt{\cosh^2(t)-1}}dt+\int \frac{\cosh(t)\sinh^3(t)}{\sqrt{\cosh^2(t)-1}}dt$$ The first half of that integral $\int \frac{\cosh(t)\sinh(t)}{\sqrt{\cosh^2(t)-1}}dt$ is extremely easy if you make the substitution $u = \cosh^2(t)-1$. For the second integral, use integration by parts with $$u = \sinh^2(t), \quad du = 2\sinh(t)\cosh(t)dt, \quad dv = \frac{\cosh(t)\sinh(t)}{\sqrt{\cosh^2(t)-1}}dt$$ where $v$ is the first half of the integral you already solved. Hence, $$\int \frac{\cosh(t)\sinh^3(t)}{\sqrt{\cosh^2(t)-1}}dt = \int \frac{\cosh(t)\sinh(t)}{\sqrt{\cosh^2(t)-1}}\sinh^2(t)dt \\ = uv-\int v du \\ = \sinh^2(t) \cdot (\text{Integral you already solved})-\int 2\sinh(t)\cosh(t)\cdot (\text{Integral you already solved})dt$$ For the new integral obtained through integration by parts, you should again be able to use $u =\cosh^2(t)-1$ for a pretty easy result.

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Let $\displaystyle u=x^2, dv=\frac{x}{\sqrt{x^2-1}}dx$ $\;\;$and $\;\;du=2xdx, v=\sqrt{x^2-1}$ to get

$\displaystyle\int\frac{x^3}{\sqrt{x^2-1}}\;dx=x^2\sqrt{x^2-1}-\int2x\sqrt{x^2-1}\;dx=x^2\sqrt{x^2-1}-\frac{2}{3}(x^2-1)^{3/2}+C$


Alternatively, let $u=\sqrt{x^2-1}, \;\;du\displaystyle=\frac{x}{\sqrt{x^2-1}}dx, \;\;x^2=u^2+1$ to get

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

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