Question

evaluate the integral.
int\frac{dx}{x^{2}sqrt{x^{2}-225}},x > 15
int\frac{dx}{x^{2}sqrt{x^{2}-225}}=square

Explanation:

Step1: Use trigonometric substitution

Let $x = 15\sec\theta$, then $dx=15\sec\theta\tan\theta d\theta$. Also, $\sqrt{x^{2}-225}=\sqrt{225\sec^{2}\theta - 225}=15\tan\theta$ and $x^{2}=225\sec^{2}\theta$.

Step2: Substitute into the integral

The integral $\int\frac{dx}{x^{2}\sqrt{x^{2}-225}}$ becomes $\int\frac{15\sec\theta\tan\theta d\theta}{225\sec^{2}\theta\times15\tan\theta}=\frac{1}{225}\int\frac{d\theta}{\sec\theta}=\frac{1}{225}\int\cos\theta d\theta$.

Step3: Integrate $\cos\theta$

$\frac{1}{225}\int\cos\theta d\theta=\frac{1}{225}\sin\theta + C$.

Step4: Express $\sin\theta$ in terms of $x$

Since $x = 15\sec\theta$, then $\sec\theta=\frac{x}{15}$ and $\cos\theta=\frac{15}{x}$. Using $\sin^{2}\theta=1 - \cos^{2}\theta$, we have $\sin\theta=\frac{\sqrt{x^{2}-225}}{x}$.

Step5: Get the final result

The integral is $\frac{\sqrt{x^{2}-225}}{225x}+C$.

Answer:

$\frac{\sqrt{x^{2}-225}}{225x}+C$