Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

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

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

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

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

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

share|improve this question
    
Note that you’re missing a $d\theta$ from every displayed line except the first and last. –  Brian M. Scott May 31 '12 at 10:44
    
Does that make any difference in the calculations? I know it is suppose to be their but I don't know if it really affects the problem. –  user138246 May 31 '12 at 10:46
    
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.) –  Brian M. Scott May 31 '12 at 10:48

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
I fixed it but I still don't know where I went wrong. –  user138246 May 31 '12 at 11:02
    
@Jordan How does $(\sqrt{5}\sin x)^2 = 5 \cos^2 x$? That's where you went wrong. –  Eugene May 31 '12 at 11:39
    
I fixed it in the problem. –  user138246 May 31 '12 at 11:41
    
@Jordan: I wouldn't say so, looking at the OP –  Ilya May 31 '12 at 11:55

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}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.