Question

Evaluate ${\int }_{-3}^1\sqrt{(x^2+6x+13)}dx$ leaving your answer in terms of natural logarithms.

Answer 1

We need to evaluate ${\int }_{-3}^1\sqrt{(x^2+6x+13)}dx$

Completing the square, $x^2+6x+13=(x+3{)}^2+4$

\since there is $+$ sign completing the square, we need to use the identity ${\sinh }^{2}A+1={\cosh }^{2}A$

Now make the substituition $x+3=2\sinh \theta$

$\Rightarrow \frac{dx}{d\theta }=2\cosh \theta$

When $x=-3,\sinh \theta =0\Rightarrow \theta =0$

When $x=1,\sinh \theta =2\Rightarrow \theta ={\sinh }^{-1}2=ln(2+\sqrt{5})$

$\therefore {\int }_{-3}^1\sqrt{(x^2+6x+13)}dx={\int }_0^{{\sinh }^{-1}2}\sqrt{4{\sinh }^2\theta +4}\cdot 2\cosh \theta d\theta$

$={\int }_0^{{\sinh }^{-1}2}4{\cosh }^2\theta d\theta$

U\sing the identity $\cosh 2A=2{\cosh }^{2}A-1$ we get

$={\int }_0^{{\sinh }^{-1}2}(2+2\cosh 2\theta )d\theta$

$={[2\theta +\sinh 2\theta ]}_0^{{\sinh }^{-1}2}$

$=2[\theta +\sinh \theta \cosh \theta {]}_0^{{\sinh }^{-1}2}$

$=2({\sinh }^{-1}2+2\sqrt{1+2^2})$

$=2ln(2+\sqrt{5})+4\sqrt{5}$