Integral of $\int \frac{du}{u \sqrt{5-u^2}}$ I am trying to find this integral and I can get the answer on wolfram of course but I do not know what is wrong with my method, having gone through it twice.
$$\int \frac{du}{u \sqrt{5-u^2}}$$
$u = \sqrt{5} \sin\theta$   and   $du= \sqrt{5} \cos \theta$
$$\int \frac{\sqrt{5} \cos \theta}{\sqrt{5} \cos \theta \sqrt{5-(\sqrt{5} \sin \theta)^2}}$$
$$\int \frac{1}{\sqrt{5-(5 \cos^2 \theta)}}$$
$$\int \frac{1}{\sqrt{5(1- \cos^2 \theta)}}$$
$$\int \frac{1}{\sqrt{5(\sin^2 \theta)}}$$
$$\frac{1}{\sqrt5}\int \frac{1}{(\sin \theta)}$$
$$\frac{1}{\sqrt5}\int \csc\theta$$
$$\frac{\ln|\csc \theta - \tan \theta|}{\sqrt5} + c$$
 A: First: don't forget to write differentials $d\theta$ - they are important and omitting them  may lead to sad mistakes. But here the problem is that you wrote $\sqrt5\cos\theta$ in the denominator for $u$ instead of $\sqrt5\sin\theta$: it is in the first row where you make the substitution - so the whole solution is incorrect although all further steps are seemed to be done in the right way.
A: You made several mistakes throughout your answer. Once corrected the answer should be
$$\int \frac{du}{u \sqrt{5-u^2}}$$
$u = \sqrt{5} \sin\theta$   and   $du= \sqrt{5} \cos \theta d\theta$
\begin{eqnarray}
\int \frac{du}{u \sqrt{5-u^2}} &=& \int\dfrac{ \sqrt{5} \cos \theta d\theta}{\sqrt{5} \sin\theta \sqrt{5 - (\sqrt{5} \sin\theta)^2}} \\
&=& \int\dfrac{  \cos \theta d\theta}{ \sin\theta \sqrt{5 - 5 \sin^2\theta}} \\
&=& \int\dfrac{  \cos \theta d\theta}{ \sin\theta \sqrt{5cos^2\theta}} \\
&=& \dfrac{1}{\sqrt{5}}\int\dfrac{  \cos \theta d\theta}{ \sin\theta \cos\theta} \\
&=& \dfrac{1}{\sqrt{5}}\int\dfrac{  d\theta}{ \sin\theta } \\
&=& \dfrac{1}{\sqrt{5}}\int \sec\theta d\theta \\
&=& \dfrac{1}{\sqrt{5}}\int \csc\theta \dfrac{ \csc \theta + \cot \theta }{ \csc \theta + \cot \theta  } d\theta\\
&=& -\dfrac{1}{\sqrt{5}} \ln|\csc\theta+\cot\theta| +C
\end{eqnarray}
Then since $\sin \theta = \dfrac{u}{\sqrt{5}}$, drawing the triangle we find out that the remaining side is $\sqrt{\sqrt{5}^2 -u^2} = \sqrt{5 -u^2}$. Therefore $\csc \theta   = \dfrac{1}{\sin \theta}= \dfrac{\sqrt{5}}{u}$ and $\cot\theta = \dfrac{1}{\tan \theta} = \dfrac{\sqrt{5 -u^2}}{u}$. So 
\begin{eqnarray}
-\dfrac{1}{\sqrt{5}} \ln|\csc\theta+\tan\theta| +C &=& -\dfrac{1}{\sqrt{5}} \ln\left|\dfrac{\sqrt{5}}{u}+\dfrac{\sqrt{5 -u^2}}{u}\right| + C \\
&=& -\dfrac{1}{\sqrt{5}}\left(-\ln|u| + \ln\left|\sqrt{5}+\sqrt{5 -u^2}\right|\right) + C
\end{eqnarray}
