Question

$\int\frac{dx}{\sqrt{9x-4x^2}}$ equals
(A) $\frac19 \ sin^{-1} \bigg\lgroup \frac{9x -8}{8}\bigg\rgroup+ C$
(B) $\frac12 \ sin^{-1} \bigg\lgroup \frac{8x -9}{9}\bigg\rgroup+ C$
(C) $\frac13 \ sin^{-1} \bigg\lgroup \frac{9x -8}{8}\bigg\rgroup+ C$
(D) $\frac12 \ sin^{-1} \bigg\lgroup \frac{9x -8}{8}\bigg\rgroup+ C$

Answer 1

Answer:

B

Step-by-step explanation:

$\displaystyle\int\frac{dx}{\sqrt{9x-4x^2}}\\ \\ \\= \tfrac12 \int\frac{dx}{\sqrt{\frac94x - x^2}}\\ \\ \\= \tfrac12 \int\frac{dx}{\sqrt{(\frac98)^2 - (x-\frac98)^2}}\\ \\ \\= \tfrac12 \sin^{-1}\left(\frac{x-\frac98}{\frac98}\right) + C\\ \\ \\= \tfrac12 \sin^{-1}\left(\frac{8x-9}{9}\right) + C$