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 $du= \sqrt{5} \cos \theta$
$$\int \frac{\sqrt{5} \cos \theta}{\sqrt{5} \cos \theta \sqrt{5-(\sqrt{5} \sin \theta)^2}}$$
$$\int \frac{1}{\sqrt{5-(5 \cos^2 \theta)}}$$
$$\int \frac{1}{\sqrt{5(1- \cos^2 \theta)}}$$
$$\int \frac{1}{\sqrt{5(\sin^2 \theta)}}$$
$$\frac{1}{\sqrt5}\int \frac{1}{(\sin \theta)}$$
$$\frac{1}{\sqrt5}\int \csc\theta$$
$$\frac{\ln|\csc \theta – \tan \theta|}{\sqrt5} + c$$
Best Answer
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.