Trig Substitution Integral

emma1

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My friends and I have each worked through this integral to come up with the same answer but it does not match the one from WolframAlpha and other online resources. We are trying to integrate: sqrt(5-x^2)/x dx. Using trig substitutions, I have come up with sqrt(5-x^2)-sqrt(5)lnabs((sqrt(5)+sqrt(5-x^2))/x)+C
This seems to be correct except the internet shows that the numerator of the insides of the natural log is off by a factor of sqrt(5). The trig substitution seems to be correct: csc= sqrt(5)/x, cot = sqrt(5-x^2)/x , cos =sqrt(5-x^2)/sqrt(5) Maybe my friends and I are missing a property of the natural log or maybe the internet is misleading us. Some clarity would be appreciated.
 
My friends and I have each worked through this integral to come up with the same answer but it does not match the one from WolframAlpha and other online resources. We are trying to integrate: sqrt(5-x^2)/x dx. Using trig substitutions, I have come up with sqrt(5-x^2)-sqrt(5)lnabs((sqrt(5)+sqrt(5-x^2))/x)+C
This seems to be correct except the internet shows that the numerator of the insides of the natural log is off by a factor of sqrt(5). The trig substitution seems to be correct: csc= sqrt(5)/x, cot = sqrt(5-x^2)/x , cos =sqrt(5-x^2)/sqrt(5) Maybe my friends and I are missing a property of the natural log or maybe the internet is misleading us. Some clarity would be appreciated.

Did you differentiate your answer (function)?

Did you get the original function back? If you did then most probably you are correct!
 
My friends and I have each worked through this integral to come up with the same answer but it does not match the one from WolframAlpha and other online resources. We are trying to integrate: sqrt(5-x^2)/x dx. Using trig substitutions, I have come up with sqrt(5-x^2)-sqrt(5)lnabs((sqrt(5)+sqrt(5-x^2))/x)+C
This seems to be correct except the internet shows that the numerator of the insides of the natural log is off by a factor of sqrt(5). The trig substitution seems to be correct: csc= sqrt(5)/x, cot = sqrt(5-x^2)/x , cos =sqrt(5-x^2)/sqrt(5) Maybe my friends and I are missing a property of the natural log or maybe the internet is misleading us. Some clarity would be appreciated.

Is your integral \(\displaystyle \displaystyle \begin{align*} \int{\frac{\sqrt{5-x^2}}{x}\,\mathrm{d}x} \end{align*}\) or \(\displaystyle \displaystyle \begin{align*} \int{\sqrt{\frac{5-x^2}{x}}\,\mathrm{d}x} \end{align*}\)?
 
My friends and I have each worked through this integral to come up with the same answer but it does not match the one from WolframAlpha and other online resources. We are trying to integrate: sqrt(5-x^2)/x dx. Using trig substitutions, I have come up with sqrt(5-x^2)-sqrt(5)lnabs((sqrt(5)+sqrt(5-x^2))/x)+C
This seems to be correct except the internet shows that the numerator of the insides of the natural log is off by a factor of sqrt(5). The trig substitution seems to be correct: csc= sqrt(5)/x, cot = sqrt(5-x^2)/x , cos =sqrt(5-x^2)/sqrt(5) Maybe my friends and I are missing a property of the natural log or maybe the internet is misleading us. Some clarity would be appreciated.
And dθ\displaystyle d\theta is?
 
Since it's \(\displaystyle \displaystyle \begin{align*} \int{ \frac{\sqrt{5-x^2}}{x}\,\mathrm{d}x } \end{align*}\) I would do...

\(\displaystyle \displaystyle \begin{align*} \int{ \frac{\sqrt{5 - x^2}}{x} \,\mathrm{d}x } &= \int{\frac{5-x^2}{x\,\sqrt{5-x^2}}\,\mathrm{d}x} \\ &= \int{ \frac{-x\,\left( 5 - x^2 \right) }{-x^2\,\sqrt{5-x^2}} \,\mathrm{d}x } \\ &= \int{ \left( \frac{5-x^2}{-x^2} \right) \left( -\frac{x}{\sqrt{5-x^2}} \right) \,\mathrm{d}x } \end{align*}\)

Now let \(\displaystyle \displaystyle \begin{align*} u = \sqrt{5-x^2} \implies \mathrm{d}u = -\frac{x}{\sqrt{5-x^2}}\,\mathrm{d}x \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} \int{ \left( \frac{5-x^2}{-x^2} \right)\left( -\frac{x}{\sqrt{5-x^2}} \right) \,\mathrm{d}x } &= \int{ \frac{u^2}{u^2 - 5} \,\mathrm{d}u } \\ &= \int{ 1 + \frac{5}{u^2 - 5} \,\mathrm{d}u } \end{align*}\)

You can now proceed with partial fractions, no trigonometric substitution necessary :)
 
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