Arc length of curve stuck with integration

Question

Data from exercise

$$y=\frac{4}{3}x^2+2\\ x\in[-1,1]$$

Formula for length of curve $$L=\int_a^{b}\sqrt{1+(f(x)')^2}\ dx$$

So far i have $$y'=\frac{8}{3}x$$

$$\int_{-1}^{1}\sqrt{1+\frac{64}{9}x^2}\ dx$$ Substition $$t^2=\frac{64}{9}x^2$$ $$t=\frac{8}{3}x$$ $$\frac{3}{8}dt=dx$$ So i have $$\int_{-\frac{8}{3}}^{\frac{8}{3}}\sqrt{1+t^2}\ dt$$

I this point i dont have a clue how to integrate this

Answer 1

Let $x = \sinh \theta \implies$ $$ \begin{align} I &= \int \sqrt{1+x^2}dx \\ &= \int \sqrt{1+\sinh^2 \theta}\cosh \theta d \theta \\ &= \int \cosh^2 \theta d \theta \\ &= \frac{1}{2}\int 1 + \cosh 2 \theta d \theta \\ &= \frac{1}{2}\theta+\frac{1}{4}\sinh 2 \theta + c\\ &= \frac{1}{2}\sinh^{-1}x+\frac{1}{2}x\sqrt{1+x^2}+c\\ &=\frac{1}{2}\ln(x + \sqrt{1+x^2})+\frac{1}{2}x\sqrt{1+x^2}+c \end{align}$$

Answer 2

Call the indefinite integral $I$

$$I\equiv\int\sqrt{1+t^2}\,\mathrm dt$$

Through integration by parts with $u=\sqrt{1+t^2}$ and $v=x$, then

$$I=t\sqrt{1+t^2}-\int\frac {t^2}{\sqrt{1+t^2}}\,\mathrm dt$$

The integrand can also be rewritten as

\begin{align*} I & =\int\frac {1+t^2}{\sqrt{1+t^2}}\,\mathrm dt\\ & =\int\frac {\mathrm dt}{\sqrt{1+t^2}}+\int\frac {t^2}{\sqrt{1+t^2}}\,\mathrm dt \end{align*}

Adding the two expressions for $I$ together and dividing both sides by two eliminates one of the integral terms.

\begin{align*} I & =\frac 12t\sqrt{1+t^2}+\frac 12\int\frac {\mathrm dt}{\sqrt{1+t^2}} \end{align*}

The last integral is somewhat well-known and can be easily evaluated with an Euler Substitution $x=t+\sqrt{1+t^2}$.

$$\int\frac {\mathrm dt}{\sqrt{1+t^2}}=\log\left(t+\sqrt{1+t^2}\right)$$

Therefore, the indefinite version of the integral becomes

$$\int\sqrt{1+t^2}\,\mathrm dt\color{blue}{=\frac 12t\sqrt{1+t^2}+\frac 12\log\left(t+\sqrt{1+t^2}\right)+C}$$

Now, substitute in the bounds for $t=8/3$ and $t=-8/3$ to arrive at your answer.

Answer 3

$$I=\int\sqrt{1+x^2} \:\:dx$$

Let $x=\operatorname{tan}(u)$ $\implies$ $dx=du \operatorname{sec}^2(u)$

therefore our integral becomes $$I=\int\operatorname{sec}^3(u)\:\:du$$ You can easily calculate this using reduction formula or by integration by parts.1

Don't forget to undo the substitution so that you can apply the limits afterwards.