3

Evaluate $$\int_3^6{{\sqrt x}\over {\sqrt x+\sqrt{9-x}}}dx $$

I understand some tricky substitution is required to work this out but I can't decide what . May be its a too easy question to be asked on this board but please give me some hint on the substitution. I can work the rest hopefully.

Martin Sleziak
  • 53,687
  • have u tried to rationalize the denominator? – tired Jan 08 '16 at 10:06
  • 3
    See http://math.stackexchange.com/questions/1073120/integral-int-12011-frac-sqrtx-sqrt2012-x-sqrtxdx – lab bhattacharjee Jan 08 '16 at 10:07
  • @AhmedHussein : ok I got $18{\sin^2 \theta \cos \theta}\over {\sin\theta + \cos \theta}$ –  Jan 08 '16 at 10:09
  • The tricky substitution you need is $u = 9-x$. – JimmyK4542 Jan 08 '16 at 10:12
  • Check out Nelsen's short paper Symmetry and Integration: http://www.maa.org/sites/default/files/pdf/mathdl/CMJ/Nelsen39-41.pdf – Hans Lundmark Jan 08 '16 at 10:15
  • @labbhattacharjee : thanks. that solved my problem . Now I find the substitution Ahmed Hussain suggested : $x=9\sin^\theta$ interesting . Can you help me a little to proceed with that now . Thanks –  Jan 08 '16 at 10:16
  • @user80631,, You will reach something like: http://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx – lab bhattacharjee Jan 08 '16 at 10:41

1 Answers1

11

Recall that $$\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$$ Hence, if $I = \displaystyle \int_3^6 \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}}dx$, we then have $$I = \int_3^6 \dfrac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{9-(9-x)}}dx = \int_3^6 \dfrac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}}dx$$ Hence, we have that $$I + I = \int_3^6 \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}}dx + \int_3^6 \dfrac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}}dx = \int_3 ^6 dx = 3$$ Hence, the integral is $\dfrac32$.

Adhvaitha
  • 20,259