Thanks!

2. Just to clarify: you're asked to compute

$\displaystyle{\int\frac{du}{u\sqrt{5-u^{2}}}}.$

Is that correct?

3. Originally Posted by softballchick

Thanks!
It would make things easier if you typed your questions out rather than using an attachment. There is nothing in your attachment that cannot be easily typed. Also, it would be easier to provide specific help if you showed what you had tried and said where you are stuck.

4. Originally Posted by softballchick

$\int_{\frac{1}{u\sqrt{5-u^2}}du$

Thanks!
One way is to draw a right-angled triangle, with perpendicular sides $u,\ \sqrt{5-u^2}$

and hypotenuse $\sqrt{5}$

then

$cos\theta=\frac{u}{\sqrt{5}}\ \Rightarrow\ u=\sqrt{5}cos\theta$

$du=-\sqrt{5}sin\theta\ d\theta$

$sin\theta=\frac{\sqrt{5-u^2}}{\sqrt{5}}\ \Rightarrow\ \sqrt{5-u^2}=\sqrt{5}sin\theta$

then use these to write a trigonometric integral

5. $\int{\frac{du}{u\sqrt{5 - u^2}}}$.

Let $u = \sqrt{5}\sin{\theta}$ so that $du = \sqrt{5}\cos{\theta}\,d\theta$ and the integral becomes

$\int{\frac{\sqrt{5}\cos{\theta}\,d\theta}{\sqrt{5} \sin{\theta}\sqrt{5 - (\sqrt{5}\sin{\theta})^2}}}$

$= \int{\frac{\cos{\theta}\,d\theta}{\sin{\theta}\sqr t{5 - 5\sin^2{\theta}}}$

$= \int{\frac{\cos{\theta}\,d\theta}{\sin{\theta}\sqr t{5(1 - \sin^2{\theta})}}$

$= \int{\frac{\cos{\theta}\,d\theta}{\sin{\theta}\sqr t{5\cos^2{\theta}}}$

$= \int{\frac{\cos{\theta}\,d\theta}{\sqrt{5}\sin{\th eta}\cos{\theta}}}$

$= \int{\frac{d\theta}{\sqrt{5}\sin{\theta}}}$

$= -\frac{\log{(\cot{\theta} + \csc{\theta})}}{\sqrt{5}} + C$

$= -\frac{\log{\left(\frac{1 + \sqrt{1 - \sin^2{\theta}}}{\sin{\theta}}\right)}}{\sqrt{5}} + C$

$= -\frac{\log{\left(\frac{1 + \sqrt{1 - \frac{u^2}{5}}}{\frac{u}{\sqrt{5}}}\right)}}{\sqrt {5}} + C$

$= -\frac{\log{\left(\frac{1 + \sqrt{\frac{5-u^2}{5}}}{\frac{u}{\sqrt{5}}}\right)}}{\sqrt{5}} + C$

$= -\frac{\log{\left(\frac{\sqrt{5} + \sqrt{5 - u^2}}{u}\right)}}{\sqrt{5}} + C$

$= -\frac{\sqrt{5}\log{\left(\frac{\sqrt{5} + \sqrt{5 - u^2}}{u}\right)}}{5} + C$