Evaluate the Integral $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$ $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$
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. 
 A: 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$$\frac{3\sqrt 3}{24}-\frac28=\frac14\left(\frac{\sqrt 3}{2}-1\right)$$
A: $$\int\frac{1}{t^3\sqrt{t^2-1}}\space\space\text{d}t=$$

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 $u=\sec^{-1}(t)$:

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

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

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

Filling in your boundaries gives us:
$$\left(\frac{\sin\left(2\sec^{-1}(2)\right)}{4}+\frac{\sec^{-1}(2)}{2}\right)-\left(\frac{\sin\left(2\sec^{-1}(\sqrt{2})\right)}{4}+\frac{\sec^{-1}(\sqrt{2})}{2}\right)=$$
$$\left(\frac{\sin\left(2\cdot\frac{\pi}{3}\right)}{4}+\frac{\frac{\pi}{3}}{2}\right)-\left(\frac{\sin\left(2\cdot\frac{\pi}{4}\right)}{4}+\frac{\frac{\pi}{4}}{2}\right)=$$
$$\left(\frac{\sin\left(\frac{2\pi}{3}\right)}{4}+\frac{\pi}{6}\right)-\left(\frac{\sin\left(\frac{\pi}{2}\right)}{4}+\frac{\pi}{8}\right)=$$
$$\left(\frac{\frac{\sqrt{3}}{2}}{4}+\frac{\pi}{6}\right)-\left(\frac{1}{4}+\frac{\pi}{8}\right)=$$
$$\left(\frac{\sqrt{3}}{8}+\frac{\pi}{6}\right)-\left(\frac{1}{4}+\frac{\pi}{8}\right)=$$
$$\frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}$$
A: 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?
A: 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 $~\dfrac{\pi+3\sqrt3-6}{12}~.$
