# How do I evaluate intsqrt(36+9x^2)dx?

Feb 15, 2015

The answer is: $\frac{3}{2} \cdot x \sqrt{4 + {x}^{2}} + 6 \arcsin h \left(\frac{x}{2}\right) + c$

First of all:

$\int \sqrt{36 + 9 {x}^{2}} \mathrm{dx} = 3 \int \sqrt{4 + {x}^{2}} \mathrm{dx}$ and now we have to substitute:

$x = 2 \sinh t \Rightarrow \mathrm{dx} = 2 \cosh t \mathrm{dt}$.

So:

$3 \int \sqrt{4 + {x}^{2}} \mathrm{dx} = 3 \int \sqrt{4 + 4 {\sinh}^{2} t} \cdot 2 \cosh t \mathrm{dt} =$

$= 3 \int 2 \sqrt{1 + {\sinh}^{2} t} \cdot 2 \cosh t \mathrm{dt} = 12 \int {\cosh}^{2} t \mathrm{dt} =$

$= 12 \int \frac{\cosh 2 t + 1}{2} \mathrm{dt} = 6 \frac{\sinh 2 t}{2} + 6 t + c =$

$= 3 \cdot 2 \sinh t \cosh t + 6 t + c =$

Now: $\sinh t = \frac{x}{2}$, $t = \arcsin h \left(\frac{x}{2}\right)$

since cosht=sqrt(1+sinh^2t, than $\cosh t = \sqrt{1 + {x}^{2} / 4} =$

$= \sqrt{\frac{4 + {x}^{2}}{4}} = \frac{1}{2} \sqrt{4 + {x}^{2}}$

So:

$I = 6 \cdot \frac{x}{2} \cdot \frac{1}{2} \sqrt{4 + {x}^{2}} + 6 \arcsin h \left(\frac{x}{2}\right) + c =$

$= \frac{3}{2} \cdot x \sqrt{4 + {x}^{2}} + 6 \arcsin h \left(\frac{x}{2}\right) + c$