evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-169}},x > 13…

evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-169}},x > 13 \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 13sin\theta)\nb. (x = 13sec\theta)\nc. (x = 13\tan\theta)\nrewrite the given integral using this substitution.\n int\frac{dx}{sqrt{x^{2}-169}}=int(sec\theta)d\theta \n(type an exact answer.)\nevaluate the integral.\n int\frac{dx}{sqrt{x^{2}-169}}=square \n(type an exact answer.)

evaluate the following integral using trigonometric substitution.\n int\frac{dx}{sqrt{x^{2}-169}},x > 13 \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 13sin\theta)\nb. (x = 13sec\theta)\nc. (x = 13\tan\theta)\nrewrite the given integral using this substitution.\n int\frac{dx}{sqrt{x^{2}-169}}=int(sec\theta)d\theta \n(type an exact answer.)\nevaluate the integral.\n int\frac{dx}{sqrt{x^{2}-169}}=square \n(type an exact answer.)

Answer

Explanation:

Step1: Apply substitution

Given $x = 13\sec\theta$, then $dx=13\sec\theta\tan\theta d\theta$. Also, $\sqrt{x^{2}-169}=\sqrt{169\sec^{2}\theta - 169}=\sqrt{169(\sec^{2}\theta - 1)} = 13\tan\theta$ (since $\sec^{2}\theta-1=\tan^{2}\theta$ and for the domain $x > 13$, $\tan\theta>0$). So, $\int\frac{dx}{\sqrt{x^{2}-169}}=\int\frac{13\sec\theta\tan\theta}{13\tan\theta}d\theta=\int\sec\theta d\theta$.

Step2: Integrate $\sec\theta$

The integral of $\sec\theta$ is $\ln|\sec\theta+\tan\theta|+C$. Since $x = 13\sec\theta$, then $\sec\theta=\frac{x}{13}$ and $\tan\theta=\frac{\sqrt{x^{2}-169}}{13}$.

Answer:

$\ln\left| \frac{x}{13}+\frac{\sqrt{x^{2}-169}}{13}\right|+C=\ln\left|x + \sqrt{x^{2}-169}\right|-\ln(13)+C=\ln\left|x+\sqrt{x^{2}-169}\right|+C_1$ (where $C_1 = C-\ln(13)$)