evaluate the integral using trigonometric substitution.\n int \frac{1}{y^{2}sqrt{15 - y^{2}}}dy \n(use…

Explanation:

Step1: Set up the substitution

Let $y = \sqrt{15}\sin\theta$, then $dy=\sqrt{15}\cos\theta d\theta$. And $\sqrt{15 - y^{2}}=\sqrt{15 - 15\sin^{2}\theta}=\sqrt{15}\cos\theta$, $y^{2}=15\sin^{2}\theta$.

Step2: Substitute into the integral

Substituting these into the integral $\int\frac{1}{y^{2}\sqrt{15 - y^{2}}}dy$, we get: [ \begin{align*} \int\frac{1}{15\sin^{2}\theta\cdot\sqrt{15}\cos\theta}\cdot\sqrt{15}\cos\theta d\theta&=\frac{1}{15}\int\frac{1}{\sin^{2}\theta}d\theta\ &=\frac{1}{15}\int\csc^{2}\theta d\theta \end{align*} ]

Step3: Integrate

We know that $\int\csc^{2}\theta d\theta=-\cot\theta + C$. Since $y = \sqrt{15}\sin\theta$, then $\sin\theta=\frac{y}{\sqrt{15}}$, and using the right - triangle relationship $\cot\theta=\frac{\sqrt{15 - y^{2}}}{y}$.

Step4: Back - substitute

So $\frac{1}{15}\int\csc^{2}\theta d\theta=-\frac{1}{15}\cdot\frac{\sqrt{15 - y^{2}}}{y}+C$.

Answer:

$-\frac{\sqrt{15 - y^{2}}}{15y}+C$