Trigonometric substitution

[nameVoid]

$$9x^2-4y^2=36$$
$$\frac{x^2}{4}-\frac{y^2}{9}=1$$
$$y=\frac{3}{2}\sqrt{x^2-4}$$
$$3\int_{2}^{3}\sqrt{x^2-4}dx$$
$$x=2sect$$
$$dx=2secttant$$
$$12\int_{a}^{b}tan^2tsectdt$$
$$12\int_{a}^{b}(sec^2t-1)(sect)dt$$
$$12\int sec^3tdt-12\int sectdt$$
$$6\int secttant-6\int ln|sect+tant|$$
$$sect=\frac{x}{2}$$
$$tant=\frac{\sqrt{x^2-4}}{2}$$
$$\frac{3x\sqrt{x^2-4}}{2}-6ln|\frac{x+\sqrt{x^2-4}}{2}| [2,3]$$
$$\frac{9\sqrt{5}}{2}-6ln|3+\sqrt{5}|+C$$

 

[vela]

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