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Question Number 62251 by hovea cw last updated on 18/Jun/19

$$\int {({\mathrm{x}}^2-4)}^{1/2}\mathrm{dx}$$$$\mathrm{trig}\mathrm{substitution}\mathrm{only}$$

Commented by maxmathsup by imad last updated on 18/Jun/19

$$letI=\int \sqrt{x^2-4}dxweusethechangementx=2ch\left(t\right)\Rightarrow$$$$I=\int \sqrt{4{ch}^{2}t-4(}2sh\left(t\right)dt=4\int \sqrt{{ch}^{2}t-1}sh\left(t\right)dt=4\int {sh}^{2}tdt$$$$=4\int \frac{ch\left(2t\right)-1}{2}dt=2\int ch\left(2t\right)dt-2t=sh\left(2t\right)-2t$$$$=2sh\left(t\right)ch\left(t\right)-2t=x\sqrt{{ch}^{2}t-1}-2t=x\sqrt{\frac{x^2}{4}-1}-2t=\frac{x}{2}\sqrt{x^2-4}-2argch\left(\frac{x}{2}\right)$$$$=\frac{x}{2}\sqrt{x^2-4}-2ln(\frac{x}{2}+\sqrt{\frac{x^2}{4}-1})=\frac{x}{2}\sqrt{x^2-4}-2ln\left(\frac{x+\sqrt{x^2-4}}{2}\right)+c_0$$$$=\frac{x}{2}\sqrt{x^2-4}-2ln(x+\sqrt{x^2-4})+4ln\left(2\right)+c_0\Rightarrow$$$$\int \sqrt{x^2-4}dx=\frac{x}{2}\sqrt{x^2-4}-2ln(x+\sqrt{x^2-4})+C.$$

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