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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 $du= \sqrt{5} \cos \theta$

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

$\int \frac{1}{\sqrt{5-(5 \cos^2 \theta)}}$

$\int \frac{1}{\sqrt{5(1- \cos^2 \theta)}}$ $\int \frac{1}{\sqrt{5(\sin^2 \theta)}}$

$\frac{1}{\sqrt5}\int \frac{1}{(\sin \theta)}$

$\frac{1}{\sqrt5}\int \csc\theta$

$\frac{\ln|\csc \theta - \tan \theta|}{\sqrt5} + c$

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    It didn’t affect the calculation in this case: the problem there is the one that Ilya noted, the incorrect substitution for $x$ in the denominator. But as he said, omitting it can lead to errors later on. (Besides, many teachers $-$ me, for one! $-$ will penalize you for omitting it.)2012-05-31

2 Answers 2

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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.

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    @Jordan: I wouldn't say so, looking at the OP2012-05-31
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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}

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    the integral of csc(x) = ln(csc x - cot x) NOT -ln(csc x + cot x). BTW, that would be ln(1/(csc x+cotx))2015-06-16