integral [(25x²-4)^1/2]/x dx

Question

kira_yamato · Answer

Do you mean integrate the following with respect to x? $$\frac{\sqrt{25x^2 - 4}}{x}$$ If this is true, then you can use integration by parts... http://en.wikipedia.org/wiki/Integration_by_parts

nikvist · Answer

$$\int\frac{\sqrt{25x^2-4}}{x}dx=\int\frac{5\sqrt{x^2-4/25}}{x}dx=5\int\sqrt{1-\left(\frac{2}{5x}\right)^2}dx$$$$\frac{2}{5x}=\cos{y}\quad,\quad x=\frac{2}{5\cos{y}}\quad,\quad dx=\frac{2\sin{y}}{5\cos^2{y}}dy$$$$5\int\sqrt{1-\left(\frac{2}{5x}\right)^2}dx=5\int\sqrt{1-\cos^2{y}}\frac{2\sin{y}}{5\cos^2{y}}dy=$$$$=2\int\frac{\sin^2{y}}{\cos^2{y}}dy=2\int\left(\frac{1}{\cos^2{y}}-1\right)dy=2(\tan{y}-y)+C=$$$$=2\left(\sqrt{\frac{1}{\cos^2{y}}-1}-y\right)+C=$$$$=2\left(\sqrt{\frac{25x^2}{4}-1}-\arccos\frac{2}{5x}\right)+C=$$$$=\sqrt{25x^2-4}-2\arccos\frac{2}{5x}+C$$