rewrite the given integral using this substitution.\n int\frac{dx}{sqrt{x^{2}-289}}=int(square)d\theta…

Explanation:

Step1: Let $x = 17\sec\theta$

Differentiate $x$ with respect to $\theta$. Since $x = 17\sec\theta$, then $dx=17\sec\theta\tan\theta d\theta$. Also, $\sqrt{x^{2}-289}=\sqrt{289\sec^{2}\theta - 289}=\sqrt{289(\sec^{2}\theta - 1)} = 17\tan\theta$ (because $\sec^{2}\theta-1=\tan^{2}\theta$ and for the appropriate domain of $\theta$, $\sqrt{\tan^{2}\theta}=\tan\theta$).

Step2: Substitute $x$ and $dx$ into the integral

Substitute $x = 17\sec\theta$, $dx = 17\sec\theta\tan\theta d\theta$ and $\sqrt{x^{2}-289}=17\tan\theta$ into $\int\frac{dx}{\sqrt{x^{2}-289}}$. We get $\int\frac{17\sec\theta\tan\theta d\theta}{17\tan\theta}$.

Step3: Simplify the integral

Cancel out the common factors $17$ and $\tan\theta$ in the integrand. The integral becomes $\int\sec\theta d\theta$.

Answer:

$\sec\theta$