Question

evaluate the integral.
int\frac{dx}{sqrt{25x^{2}-81}},x > \frac{9}{5}
which substitution transforms the given integral into one that can be solved?
a. (x=\frac{9}{5}sec\theta)
b. (x = \frac{9}{5}sin\theta)
c. (x=\frac{9}{5}\tan\theta)

Explanation:

Step1: Recall the form of substitution

For integrals of the form $\int\frac{dx}{\sqrt{a^{2}x^{2}-b^{2}}}$, we use the substitution $ax = b\sec\theta$. Here $a = 5$ and $b=9$, so $5x=9\sec\theta$ or $x=\frac{9}{5}\sec\theta$.

Step2: Analyze the substitution effect

If $x = \frac{9}{5}\sec\theta$, then $dx=\frac{9}{5}\sec\theta\tan\theta d\theta$ and $\sqrt{25x^{2}-81}=\sqrt{81\sec^{2}\theta - 81}=9\tan\theta$ (since $\sec^{2}\theta-1=\tan^{2}\theta$ and for the domain $x>\frac{9}{5}$, $\tan\theta>0$). The integral $\int\frac{dx}{\sqrt{25x^{2}-81}}$ becomes $\int\frac{\frac{9}{5}\sec\theta\tan\theta d\theta}{9\tan\theta}=\frac{1}{5}\int\sec\theta d\theta$ which is a standard - form integral.

Answer:

A. $x = \frac{9}{5}\sec\theta$