F femmed0ll New member Joined Aug 9, 2010 Messages 14 Aug 9, 2010 #1 4)? ( x?x + ?x ) / (x^2) dx 5)? (sec?) / (sec ? - cos?) d? we have a test on Tuesday, and i have no idea what steps to do to get this answer! thanks
4)? ( x?x + ?x ) / (x^2) dx 5)? (sec?) / (sec ? - cos?) d? we have a test on Tuesday, and i have no idea what steps to do to get this answer! thanks
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Aug 9, 2010 #2 Hello, femmed0ll! \(\displaystyle 4)\;\int \frac{x\sqrt{x} + \sqrt{x}}{x^2}\,dx\) Click to expand... How about a little Algebra? \(\displaystyle \int\frac{x^{\frac{3}{2}} + x^{\frac{1}{2}}}{x^2}\,dx \;=\;\int \left(x^{-\frac{1}{2}} + x^{-\frac{3}{2}}\right)\,dx \;=\;2x^{\frac{1}{2}} - 2x^{-\frac{1}{2}} + C\) \(\displaystyle 5)\;\int \frac{\sec\theta}{\sec\theta - \cos\theta}\,d\theta\) Click to expand... \(\displaystyle \text{We have: }\;\int\frac{\frac{1}{\cos\theta}}{\frac{1}{\cos\theta} - \cos\theta}\,d\theta\) \(\displaystyle \text{Multiply by }\frac{\cos\theta}{\cos\theta}\!:\;\;\int\frac{1}{1-\cos^2\!\theta}\,d\theta \;=\;\int\frac{d\theta}{\sin^2\!\theta} \;=\;\int\csc^2\!\theta\,d\theta \;=\;-\cot\theta + C\)
Hello, femmed0ll! \(\displaystyle 4)\;\int \frac{x\sqrt{x} + \sqrt{x}}{x^2}\,dx\) Click to expand... How about a little Algebra? \(\displaystyle \int\frac{x^{\frac{3}{2}} + x^{\frac{1}{2}}}{x^2}\,dx \;=\;\int \left(x^{-\frac{1}{2}} + x^{-\frac{3}{2}}\right)\,dx \;=\;2x^{\frac{1}{2}} - 2x^{-\frac{1}{2}} + C\) \(\displaystyle 5)\;\int \frac{\sec\theta}{\sec\theta - \cos\theta}\,d\theta\) Click to expand... \(\displaystyle \text{We have: }\;\int\frac{\frac{1}{\cos\theta}}{\frac{1}{\cos\theta} - \cos\theta}\,d\theta\) \(\displaystyle \text{Multiply by }\frac{\cos\theta}{\cos\theta}\!:\;\;\int\frac{1}{1-\cos^2\!\theta}\,d\theta \;=\;\int\frac{d\theta}{\sin^2\!\theta} \;=\;\int\csc^2\!\theta\,d\theta \;=\;-\cot\theta + C\)
