Evaluate the Integral $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$

Question

$\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$

I believe I've done everything right; however, my answer does not resemble the answer in the book. I think it has something to do with my Algebra. Please tell me what I am doing wrong.

Answer 1

You answer is correct indeed. The integral is

$$\begin{align}\int_{\sqrt 2}^2\frac{1}{t^3\sqrt{t^2-1}}\,dt&=\int_{\pi/4}^{\pi/3} \cos^2 x\,dx\\\\ &=\left.\left(\frac12+\frac14\sin 2x\right)\right|_{\pi/4}^{\pi/3}\\\\ &=\frac{\pi}{24}+\frac14\left(\frac{\sqrt 3}{2}-1\right) \end{align}$$

which is the same as your answer since $$\frac{4\pi}{24}-\frac\pi8=\frac{\pi}{24}$$and$$\frac{3\sqrt 3}{24}-\frac28=\frac14\left(\frac{\sqrt 3}{2}-1\right)$$

Answer 2

$$\int\frac{1}{t^3\sqrt{t^2-1}}\space\space\text{d}t=$$

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Substitute $t=\sec(u)$ and $\text{d}t=\tan(u)\sec(u)\space\space\text{d}u$. Then $\sqrt{t^2-1}=\sqrt{\sec^2(u)-1}=\tan(u)$ and $u=\sec^{-1}(t)$:

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$$\int\cos^2(u)\space\space\text{d}u=$$ $$\int\left(\frac{1}{2}\cos(2u)+\frac{1}{2}\right)\space\space\text{d}u=$$ $$\frac{1}{2}\int\cos(2u)\space\space\text{d}u+\frac{1}{2}\int 1\space\space\text{d}u=$$

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Substitute $s=2u$ and $\text{d}s=2\space\space\text{d}u$:

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$$\frac{1}{4}\int\cos(s)\space\space\text{d}s+\frac{1}{2}\int 1\space\space\text{d}u=$$ $$\frac{1}{4}\sin(s)+\frac{u}{2}+\text{C}=$$ $$\frac{\sin\left(s\right)}{4}+\frac{u}{2}+\text{C}=$$ $$\frac{\sin\left(2u\right)}{4}+\frac{u}{2}+\text{C}=$$ $$\frac{\sin\left(2\sec^{-1}(t)\right)}{4}+\frac{\sec^{-1}(t)}{2}+\text{C}$$

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Filling in your boundaries gives us:

$$\left(\frac{\sin\left(2\sec^{-1}(2)\right)}{4}+\frac{\sec^{-1}(2)}{2}\right)-\left(\frac{\sin\left(2\sec^{-1}(\sqrt{2})\right)}{4}+\frac{\sec^{-1}(\sqrt{2})}{2}\right)=$$ $$\left(\frac{\sin\left(2\cdot\frac{\pi}{3}\right)}{4}+\frac{\frac{\pi}{3}}{2}\right)-\left(\frac{\sin\left(2\cdot\frac{\pi}{4}\right)}{4}+\frac{\frac{\pi}{4}}{2}\right)=$$ $$\left(\frac{\sin\left(\frac{2\pi}{3}\right)}{4}+\frac{\pi}{6}\right)-\left(\frac{\sin\left(\frac{\pi}{2}\right)}{4}+\frac{\pi}{8}\right)=$$ $$\left(\frac{\frac{\sqrt{3}}{2}}{4}+\frac{\pi}{6}\right)-\left(\frac{1}{4}+\frac{\pi}{8}\right)=$$ $$\left(\frac{\sqrt{3}}{8}+\frac{\pi}{6}\right)-\left(\frac{1}{4}+\frac{\pi}{8}\right)=$$ $$\frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}$$

Answer 3

Your integral is well calculated.

Computing the last part for a prominent answer, we get $$\frac{1}{2}[1+\frac{1}{2}\sin 2\theta]^{\frac{\pi}{3}}_{\frac{\pi}{4}}$$ $$=\frac{1}{2}[(\frac{\pi}{3}-\frac{\pi}{4})+\frac{1}{2}(\sin (\frac{2\pi}{3})-\sin (\frac{2\pi}{4}))]$$ $$=\frac{1}{2}[\frac{\pi}{12}+\frac{1}{2}(\frac{\sqrt{3}}{2}-1)]$$ $$=\frac{\pi}{24}+\frac{1}{8}(\sqrt{3}-2)$$ Does it match the answer?

Answer 4

You did nothing wrong. Here are two alternate approaches:

$$\int\frac{dt}{t^3\sqrt{t^2-1}}=\int\frac{t}{t^4\sqrt{t^2-1}}~dt=\frac12\int\frac{d\big(t^2-1\big)}{\Big[\big(t^2-1\big)+1\Big]^2\sqrt{t^2-1}}=\frac12\int\frac{du}{\big(u+1\big)^2\sqrt u}$$

$$=\frac12\int\frac{d\big(v^2\big)}{\big(v^2+1\big)^2~v}=\int\frac{dv}{\big(v^2+1\big)^2}$$

Or, substituting $t=\cosh x$, and using $\cosh'x=\sqrt{\cosh^2x-1}=\sinh x$, we have

$$\int\frac{\sinh x}{\cosh^3x~\sinh x}~dx=\int\frac{\cosh x}{\cosh^4x}~dx=\int\frac{d\big(\sinh x\big)}{\big(1+\sinh^2x\big)^2}=\int\frac{dv}{\big(v^2+1\big)^2}$$

Using the fact that $~\dfrac1{\big(v^2+1\big)^2}=-\bigg[\dfrac{d}{da}\bigg(\dfrac1{v^2+a}\bigg)\bigg]_{a=1}$ and $~\displaystyle\int\frac{dv}{v^2+a}=\frac1{\sqrt a}~\arctan\frac v{\sqrt a}$

we soon arrive at the desired conclusion, $I=-\bigg[\dfrac{d}{da}~\bigg(\dfrac1{\sqrt a}~\arctan\dfrac v{\sqrt a}\bigg)\bigg]_{a=1}=~\dfrac{\arctan v}2~+$

$+~\dfrac12\cdot\dfrac{v}{v^2+1},~$ which, when evaluated between $v=\sqrt3$ and $v=1$, yields $~\dfrac{\pi+3\sqrt3-6}{12}~.$