Question

evaluate the integral.
int\frac{sqrt{y^{2}-49}}{y}dy,y > 7

int\frac{sqrt{y^{2}-49}}{y}dy=square

Explanation:

Step1: Use trig - substitution

Let $y = 7\sec\theta$, then $dy=7\sec\theta\tan\theta d\theta$. Since $y > 7$, $\theta\in(0,\frac{\pi}{2})$. Also, $\sqrt{y^{2}-49}=\sqrt{49\sec^{2}\theta - 49}=7\tan\theta$.

Step2: Rewrite the integral

The integral $\int\frac{\sqrt{y^{2}-49}}{y}dy$ becomes $\int\frac{7\tan\theta}{7\sec\theta}\cdot7\sec\theta\tan\theta d\theta=\int7\tan^{2}\theta d\theta$.

Step3: Use trig - identity

Recall that $\tan^{2}\theta=\sec^{2}\theta - 1$. So, $\int7\tan^{2}\theta d\theta = 7\int(\sec^{2}\theta - 1)d\theta$.

Step4: Integrate term - by - term

$7\int(\sec^{2}\theta - 1)d\theta=7(\tan\theta-\theta)+C$.

Step5: Back - substitute

Since $y = 7\sec\theta$, $\sec\theta=\frac{y}{7}$ and $\tan\theta=\frac{\sqrt{y^{2}-49}}{7}$, and $\theta=\text{arcsec}(\frac{y}{7})$.
So the integral is $\sqrt{y^{2}-49}-7\text{arcsec}(\frac{y}{7})+C$.

Answer:

$\sqrt{y^{2}-49}-7\text{arcsec}(\frac{y}{7})+C$