How do I evaluate $\int \sqrt{36 + 9 x^{2}} d x$ $intsqrt(36+9x^2)dx$ ?

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Answer 1 · 2015-02-15T19:16:47

The answer is: $\frac{3}{2} \cdot x \sqrt{4 + x^{2}} + 6 arcsin h (\frac{x}{2}) + c$ $3/2*xsqrt(4+x^2)+6arcsinh(x/2)+c$

First of all:

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

$x = 2 sinh t \Rightarrow d x = 2 cosh t d t$ $x=2sinhtrArrdx=2coshtdt$ .

So:

$3 \int \sqrt{4 + x^{2}} d x = 3 \int \sqrt{4 + 4 {sinh}^{2} t} \cdot 2 cosh t d t =$ $3intsqrt(4+x^2)dx=3intsqrt(4+4sinh^2t)*2coshtdt=$

$= 3 \int 2 \sqrt{1 + {sinh}^{2} t} \cdot 2 cosh t d t = 12 \int {cosh}^{2} t d t =$ $=3int2sqrt(1+sinh^2t)*2coshtdt=12intcosh^2tdt=$

$= 12 \int \frac{cosh 2 t + 1}{2} d t = 6 \frac{sinh 2 t}{2} + 6 t + c =$ $=12int(cosh2t+1)/2dt=6(sinh2t)/2+6t+c=$

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

Now: $sinh t = \frac{x}{2}$ $sinht=x/2$ , $t = arcsin h (\frac{x}{2})$ $t=arcsinh(x/2)$

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

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

So:

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

$= \frac{3}{2} \cdot x \sqrt{4 + x^{2}} + 6 arcsin h (\frac{x}{2}) + c$ $=3/2*xsqrt(4+x^2)+6arcsinh(x/2)+c$

How do I evaluate ∫√36+9x2dxintsqrt(36+9x^2)dx?