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I have tried calculating the integral:

$$ \int_0^a\frac{\sqrt x}{\sqrt x + \sqrt{a-x}}\mathrm{d}x $$

but getting rid of the square roots from the problem is like impossible for me.

hardmath
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Wilmot Gray
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  • Take carefully to my editing: is that what you meant to write? Be careful... – DonAntonio Jun 14 '17 at 14:45
  • yes thanks alot – Wilmot Gray Jun 14 '17 at 14:45
  • Have you tried the substitution $x=a,\sin^2(u)$? – Mark Viola Jun 14 '17 at 14:51
  • Substitute $x\to x-a$ and then add the resulting integral to the initial one. Good luck – Bumblebee Jun 14 '17 at 14:51
  • Set $x/a=\sin^2t$ and use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx/439856#439856 – lab bhattacharjee Jun 14 '17 at 16:14
  • Hi, Doris, you asked this yesterday: https://math.stackexchange.com/questions/2321871/please-integrate-int-frac-sqrtx-sqrtx-sqrta-x-dx – amWhy Jun 14 '17 at 16:29
  • Duplicate of question, asked yesterday: https://math.stackexchange.com/questions/2321871/please-integrate-int-frac-sqrtx-sqrtx-sqrta-x-dx – amWhy Jun 14 '17 at 16:43
  • The duplicate target is not quite a duplicate, because this time we have a definite integral, and can use the obvious symmetry. That trick has been seen here enough many times already, so I'm all for closing. But, the askers are either housemates or aliases, so I have to ask. Wilmot, why didn's you edit the first version into shape. Why reask? – Jyrki Lahtonen Jun 14 '17 at 19:56
  • because i do not know who ask the question first – Wilmot Gray Jun 14 '17 at 20:58

3 Answers3

7

Let $u=a-x$. Then

$$\int_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx=\int_a^0\frac{-\sqrt{a-u}}{\sqrt{a-u}+\sqrt{u}}du=\int_0^a\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx$$

So

$$2\int_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx=\int_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx+\int_0^a\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx=\int_0^adx$$

CY Aries
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  • How did you go from the second expression to the third expression in the first line ($a-u \rightarrow a-x$) – Χpẘ Jun 14 '17 at 15:35
  • $u$ and $x$ are dummy variables. The integrals do not depend on them. If $F(x)=\int\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx$, then $F(u)=\int\frac{\sqrt{a-u}}{\sqrt{u}+\sqrt{a-u}}du$. So, $\int_0^a\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx=F(a)-F(0)$ and $\int_0^a\frac{\sqrt{a-u}}{\sqrt{u}+\sqrt{a-u}}du=F(a)-F(0)$. They are the same and we can simply rename $u$ by $x$. – CY Aries Jun 14 '17 at 15:39
  • makes sense - just a simple renaming, not a substitution. I also just noticed that the values being integrated in first and third expressions are reflections around $x=a/2$, so the integrals from 0 to $a$ must be equal. – Χpẘ Jun 14 '17 at 15:55
  • Yes, that's why we choose this substitution. – CY Aries Jun 14 '17 at 15:56
3

Try exploiting the fact that $$\int_0^af(x)=\int_0^af(a-x)$$

kingW3
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1

Hint: write your Integrand in the form $$\frac{1}{1+\sqrt{\frac{a-x}{x}}}$$ and set $$t=\sqrt{\frac{a-x}{x}}$$ if you do that then you will get the indefinite integral $$-a\int\frac{2t}{(1+t)(t^2+1)}dt$$

Dr. Sonnhard Graubner
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