Question

Help with cal 2 exercise involving integrals?

Answer 1

$\displaystyle\int\frac{\sqrt{x^2+1}}x\,\mathrm dx=\int\frac{\sqrt{\tan^2B+1}}{\tan B}\sec^2B\,\mathrm dB$

$\tan^2B+1=\sec^2B$

$\sqrt{\sec^2B}=\sec B$ (provided that $\sec B>0$)

Then

$\dfrac{\sec B}{\tan B}\cdot\sec^2B=\dfrac{\sec^2B}{\tan B}\cdot\sec B$

(just moving around a factor of $\sec B$)

$\dfrac{\sec B}{\tan B}\cdot\sec^2B=\dfrac{\sec^2B}{\tan^2B}\cdot\sec B\tan B$

(multiply by $\dfrac{\tan B}{\tan B}$)