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$$ \int \sqrt{a- x \over x} dx$$


Substituting $u = \sqrt{a - x}$,

$$\int \sqrt{a- x \over x} dx = -2\int {u^2 \over \sqrt{a - u^2}} du$$

Now, for $\sqrt{a}\sin t = u$,

$$-2\int {a\sin^2 t \over \sqrt{a} \cos t} \ \ dt \ \ \sqrt{a }\cos t = -a\int(1 - \cos 2t) dt = a(t - \cos t \sin t) + C$$

Substituting for $\displaystyle t = \arcsin\left({u\over \sqrt{a}}\right) =\arcsin\left({\sqrt{a - x}\over \sqrt{a}}\right) $ I get the answer as

$$\bbox[7px,Border:2px solid black]{a\left(\sqrt{x\over a} \sqrt{a -x \over a}- \arcsin\sqrt{a -x \over a}\right) + C}$$,

But the given answer is $\displaystyle {a\left(\sqrt{x\over a} \sqrt{a -x \over a}+ \arcsin\sqrt{x \over a}\right) + C}$

I am sure that $\displaystyle \arcsin\sqrt{x \over a} \ne - \arcsin\sqrt{a -x \over a}$.

Where did I got wrong ?

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You didn't go wrong - actually, we have that $$\arcsin\sqrt{x \over a} = \frac{\pi}{2} - \arcsin\sqrt{a -x \over a}$$ as $$\left(\sqrt{x \over a} \right)^2 + \left(\sqrt{a -x \over a} \right)^2=1.$$

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  • $\begingroup$ I should really learn inverse trig identities. Even WA got it wrong, wolframalpha.com/input/…. $\endgroup$ – A---B Apr 18 '17 at 18:08
  • $\begingroup$ I think a better check is to differentiate the primitive you get - this seems less likely to give ambiguous answers. It also circumvents the whole $``+C"$ business. $\endgroup$ – πr8 Apr 18 '17 at 18:26
  • $\begingroup$ Yes thank you very much. $\endgroup$ – A---B Apr 18 '17 at 18:32

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