Question

evaluate the integral. (int\frac{sqrt{y^{2}-9}}{y}dy,y > 3) a. (y = 3sec\theta) b. (y = 3sin\theta) c. (y = 3\tan\theta) given the expression for y above, find dy in terms of (\theta). (dy=3sec\theta\tan\theta d\theta) evaluate the integral. (int\frac{sqrt{y^{2}-9}}{y}dy=)

Explanation:

Step1: Substitute $y = 3\sec\theta$

Since $y = 3\sec\theta$, then $\sqrt{y^{2}-9}=\sqrt{9\sec^{2}\theta - 9}=3\tan\theta$ (because $\sec^{2}\theta-1=\tan^{2}\theta$ and for $y>3$, $\tan\theta>0$). Also, $dy = 3\sec\theta\tan\theta d\theta$. The integral $\int\frac{\sqrt{y^{2}-9}}{y}dy$ becomes $\int\frac{3\tan\theta}{3\sec\theta}\cdot3\sec\theta\tan\theta d\theta=\int3\tan^{2}\theta d\theta$.

Step2: Rewrite $\tan^{2}\theta$

Recall the identity $\tan^{2}\theta=\sec^{2}\theta - 1$. So, $\int3\tan^{2}\theta d\theta=3\int(\sec^{2}\theta - 1)d\theta$.

Step3: Integrate term - by - term

We know that $\int\sec^{2}\theta d\theta=\tan\theta$ and $\int1d\theta=\theta$. So, $3\int(\sec^{2}\theta - 1)d\theta=3(\tan\theta-\theta)+C$.

Step4: Back - substitute

Since $y = 3\sec\theta$, then $\sec\theta=\frac{y}{3}$ and $\tan\theta=\sqrt{\sec^{2}\theta - 1}=\sqrt{\frac{y^{2}}{9}-1}=\frac{\sqrt{y^{2}-9}}{3}$, and $\theta=\text{arcsec}(\frac{y}{3})$.
The integral $\int\frac{\sqrt{y^{2}-9}}{y}dy=\sqrt{y^{2}-9}-3\text{arcsec}(\frac{y}{3})+C$.

Answer:

$\sqrt{y^{2}-9}-3\text{arcsec}(\frac{y}{3})+C$