Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.6 Exercises - Page 555: 7


The result is $$\int\frac{1}{\sqrt x(1-\cos\sqrt x)}dx=-2\cot\frac{\sqrt x}{2}+c.$$

Work Step by Step

To solve the integral $$\int\frac{1}{\sqrt x(1-\cos\sqrt x)}dx$$ we will use substitution $\sqrt x=t$ which gives us $\frac{1}{2\sqrt x}dx=dt\Rightarrow \frac{dx}{\sqrt x}=2dt. $ Putting this into the integral we get: $$\int\frac{1}{\sqrt x(1-\cos\sqrt x)}dx=\int\frac{2dt}{1-\cos t}.$$ Now we will use substitution $\tan\frac{t}{2}=z$ which gives us $\cos t=\frac{1-z^2}{1+z^2},dt=\frac{2}{1+z^2}dz.$ Substituting this into the integral we have: $$\int\frac{2dt}{1-\cos t}=\int\frac{2}{1-\frac{1-z^2}{1+z^2}}\frac{2}{1+z^2}dz=\int\frac{4}{\frac{1+z^2-1+z^2}{1+z^2}}\frac{1}{1+z^2}dz=4\int\frac{1+z^2}{2z^2}\frac{1}{1+z^2}dz=2\int\frac{dz}{z^2}=2\frac{-1}{z}+c,$$ where $c$ is arbitrary constant. Now we have to express the solution in terms of $x$ and to do that we will express $z$ in terms of $t$ and then $t$ in terms of $x$: $$\int\frac{1}{\sqrt x(1-\cos\sqrt x)}dx=-\frac{2}{z}+c=-\frac{2}{\tan\frac{t}{2}}+c=-2\cot\frac{t}{2}+c=-2\cot\frac{\sqrt x}{2}+c.$$
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