-1
$\begingroup$

Evaluate

$$\int_{-a}^{a} \sqrt\frac{a-x}{a+x}\,\mathrm{d}x.$$

answer option

  • a
  • $πa$
  • $\frac{a}{2}$
  • 0
    Question is unreadable. – 2017-02-10
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    The OP's last post was just as unreadable – 2017-02-10
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    @BrevanEllefsen I am not able to write – 2017-02-10
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    For $\sqrt{\dfrac{a-x}{a+x}}$ put $x=a\cos2t$ – 2017-02-10
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    Who edited it ? Thank you – 2017-02-10
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    Welcome to MSE, you have our [tutorial](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) and also with mouse right click: Show Math As > TeX Commands you can explore formulas.Good luck. – 2017-02-10
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    @BrevanEllefsen Thanks for your rollback. – 2017-02-10
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    @GNUSupporter No problem. The way I see it is if the OP wants to edit the corrected MathJax then go ahead, but what is up now is way better than before. The OP did the same thing on their last post around an hour ago, and undid my edit. I was annoyed enough I wasn't about to let it happen again to someone else ;) – 2017-02-10
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    this question is still open after 5 hours? wtf? – 2017-02-10
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    @5xum Now question is edited pls help to solve this ... – 2017-02-11

1 Answers 1

3

$$\int_{-a}^{a} \sqrt\frac{a-x}{a+x}\,\mathrm{d}x$$ We apply the substitution $x\to ax$ $$=a\int_{-1}^{1} \sqrt\frac{1-x}{1+x}\,\mathrm{d}x$$
Shifting horizontally a bit we get $$=a\int_{0}^{2}\sqrt{\frac{1-\left(x-1\right)}{1+\left(x-1\right)}}\mathrm{d}x$$
We can now solve the inverse of the function we are integrating for $y$ $$=2a\int_0^\infty\frac{dx}{x^2+1}=\pi a$$