evaluate the integral. (int\frac{dx}{sqrt{25x^{2}-81}},x > \frac{9}{5}) (x=\frac{9}{5}sec\theta) (x =…

Explanation:

Step1: Choose substitution

We use the substitution (x = \frac{9}{5}\sec\theta) (since for (\int\frac{dx}{\sqrt{a^{2}x^{2}-b^{2}}}), the substitution (x=\frac{b}{a}\sec\theta) is appropriate, here (a = 5), (b=9)). Then (dx=\frac{9}{5}\sec\theta\tan\theta d\theta).

Step2: Rewrite the integrand

Substitute (x) into the integral (\int\frac{dx}{\sqrt{25x^{2}-81}}). First, (25x^{2}-81=25\times\frac{81}{25}\sec^{2}\theta - 81=81\sec^{2}\theta - 81 = 81(\sec^{2}\theta - 1)=81\tan^{2}\theta) (using the identity (\sec^{2}\theta-1=\tan^{2}\theta)). So (\sqrt{25x^{2}-81}=9\tan\theta) (since (x>\frac{9}{5}), (\theta\in(0,\frac{\pi}{2})) and (\tan\theta>0)). The integral becomes (\int\frac{\frac{9}{5}\sec\theta\tan\theta d\theta}{9\tan\theta}=\frac{1}{5}\int\sec\theta d\theta).

Step3: Integrate

The antiderivative of (\sec\theta) is (\ln|\sec\theta+\tan\theta|+C). So (\frac{1}{5}\int\sec\theta d\theta=\frac{1}{5}\ln|\sec\theta+\tan\theta|+C).

Step4: Back - substitute

Since (x = \frac{9}{5}\sec\theta), then (\sec\theta=\frac{5x}{9}). Using (\tan^{2}\theta=\sec^{2}\theta - 1), we have (\tan\theta=\sqrt{\frac{25x^{2}}{81}-1}=\frac{\sqrt{25x^{2}-81}}{9}). So the integral is (\frac{1}{5}\ln\left|\frac{5x}{9}+\frac{\sqrt{25x^{2}-81}}{9}\right|+C=\frac{1}{5}\ln|5x + \sqrt{25x^{2}-81}|-\frac{1}{5}\ln9 + C=\frac{1}{5}\ln|5x+\sqrt{25x^{2}-81}|+C_1) (where (C_1 = C-\frac{1}{5}\ln9)).

Answer:

(\frac{1}{5}\ln|5x+\sqrt{25x^{2}-81}|+C)