## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{1}{10} \sec^{-1} (\dfrac{y}{5})-\dfrac{\sqrt {y^2-25}}{2y^2}+C$
Let us consider $y = 5 \sec \theta \implies dy=5 (\sec\theta) (\tan \theta) d \theta$ Now, the given integral becomes: $\int \dfrac{\sqrt{(25\sec^2 \theta-25)}(5 \sec \theta \tan \theta)}{125 \sec^3 \theta} d \theta = \int (\dfrac{1}{5}) \dfrac{\sec^2 \theta-1}{\sec^2 \theta } d \theta$ As we know that $\sec^2 \theta-1 =\tan^2 \theta$ and $\cos^2 \theta =(1/2)+(1/2) \cos 2 \theta$ or, $\dfrac{1}{5} \int (1-\cos^2 \theta) d \theta =\dfrac{1}{5} \int (1-\dfrac{1}{2} -\dfrac{1}{2} \cos ( 2\theta ))d \theta$ Now, integrate $\dfrac{1}{5}( \dfrac{1}{2} \theta -\dfrac{1}{4} \sin (2\theta))+C= \dfrac{1}{10} \theta -\dfrac{1}{20} (2 \sin \theta \cos \theta ))+C$ Plug in $\sec \theta =(y/5) \implies \theta =sec^{-1} (y/5)$ and and $\sec \theta =\dfrac{y}{5} \implies \cos \theta =(5/y)$ and $\sin \theta =\dfrac{\sqrt {y^2-25}}{y}$ : $=\dfrac{1}{10} \sec^{-1} (\dfrac{y}{5})-\dfrac{1}{10} (\dfrac{\sqrt {y^2-25}}{y}) (\dfrac{5}{y})+C$ or, $= \dfrac{1}{10} \sec^{-1} (\dfrac{y}{5})-\dfrac{\sqrt {y^2-25}}{2y^2}+C$