Integral of $\sqrt{2+t^2}$

Question

How can I solve the integral of $$ \int \sqrt{2+t^2}\,dt$$ I obtained it by calculating the length of the arc $a(t)=(t\cos t,t \sin t,t)$ with $t \in [0,2\pi]$.

I try with substitution with the function $\sinh$ but i don't manage to resolve it.

Answer 1

Start by factoring out $\sqrt2$, then some substitution:

$$\begin{align}\int\sqrt{2+t^2}\ dt&=\sqrt2\int\sqrt{1+(t/\sqrt2)^2}\ dt\\&=2\int\sqrt{1+\tan^2u}\sec^2u\ du\\&=2\int\sec^3u\ du\end{align}$$

This integral is thus well known and has its own Wikipedia page and may be done with various methods such as integration by parts.

Answer 2

Hint $$\sqrt{2+t^2}=\sqrt{2}\cdot \sqrt{1+\left(\frac{t}{\sqrt{2}}\right)^2}$$

now make:

$$\frac{t}{\sqrt{2}}=\tan x$$

Answer 3

The $\sinh$ change works when the radicand is $t^2+1$.

When instead of $1$ you have another positive number $a$, you can do this change before: $$t=\sqrt a u, dt=\sqrt adu$$ to get $$\int\sqrt{a+t^2}dt=\int a\sqrt{1+u^2}du$$

Answer 4

One way is to integrate by parts: $$ \int_0^{2\pi} \underbrace{\sqrt{t^2 + 2}}_u \, \underbrace{dt}_{dv} = \underbrace{\sqrt{t^2 + 2}}_u \cdot \underbrace{t}_v \bigg|_0^{2\pi} - \int_0^{2\pi} \underbrace{t}_v \cdot \underbrace{\frac t{\sqrt{t^2+2}} \, dt}_{du} $$ Continuing without the labels: \begin{align*} I &= \int_0^{2\pi} \sqrt{t^2 + 2} \, dt\\[0.3cm] &= t\sqrt{t^2 + 2} \bigg|_0^{2\pi} - \int_0^{2\pi} \frac{t^2}{\sqrt{t^2+2}} \, dt\\[0.3cm] &= 2\pi \sqrt{4\pi^2+2} - \int_0^{2\pi} \frac{t^2 \color{red}{+2-2}}{\sqrt{t^2+2}} \, dt\\[0.3cm] &= 2\pi \sqrt{4\pi^2+2} - \left[\int_0^{2\pi} \frac{t^2 + 2}{\sqrt{t^2+2}} \, dt - \int_0^{2\pi} \frac 2{\sqrt{t^2+2}} \, dt\right]\\[0.3cm] &= 2\pi \sqrt{4\pi^2+2} - \underbrace{\int_0^{2\pi} \sqrt{t^2+2} \, dt}_{\text{this is $I$}} + \int_0^{2\pi} \frac 2{\sqrt{t^2+2}} \, dt\\[0.3cm] &= 2\pi \sqrt{4\pi^2+2} - I + 2 \sinh^{-1}\left(\frac t{\sqrt2}\right)\bigg|_0^{2\pi}\\[0.3cm] &= 2\pi \sqrt{4\pi^2+2} - I + 2\sinh^{-1}(\pi\sqrt2) \end{align*} Therefore $$ I = 2\pi \sqrt{4\pi^2+2} - I + 2\sinh^{-1}(\pi\sqrt2). $$ Solve for $I$ to get $$ I = \pi \sqrt{4\pi^2+2} + \sinh^{-1}(\pi\sqrt2). $$

Admittedly this does require knowledge of that inverse hyperbolic sine integral (although there are other ways to evaluate that one), which may make this method not that helpful, but here it is for the sake of variety.