anonymous
  • anonymous
Evaluate the indefinite integral (2+x)/(sqrt(1-x^2)) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1370823001916:dw|
anonymous
  • anonymous
|dw:1370823179144:dw|
anonymous
  • anonymous
When you do one of the differentiations within the substitution you end up with an extra 2 which would cancel out the 1/2

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anonymous
  • anonymous
Oops I didn't see that for some reason. Good catch!
Jhannybean
  • Jhannybean
\[\large \int\limits \frac{(2+x)}{\sqrt{1-x^2}}dx\]\[\large \int\limits \frac{2}{\sqrt{1-x^2}}dx+\int\limits \frac{x}{\sqrt{1-x^2}}dx\]\[\large u=\sqrt{1-x^2}\]\[du=\frac{-x}{\sqrt{1-x^2}}dx \implies \frac{\sqrt{1-x^2}}{-x}du=dx \implies dx =\frac{u}{-\sqrt{1-u^2}}du\] solving for the first part \[\large \int\limits \frac{2}{\sqrt{1-x^2}}dx\]\[\large 2 \int\limits\limits \frac{1}{\cancel u}*(-\frac{\cancel u}{\sqrt{1-u^2}})du\]\[\large -2 \int\limits \frac{1}{\sqrt{1-u^2}}du = -2\sin^{-1}(u)+c = -2\sin^{-1}(\sqrt{1-x^2})+c\] solving for the second part\[\large \int\limits \frac{x}{\sqrt{1-x^2}}dx\]\[\large u= 1-x^2 \]\[\large du= -2xdx \implies \frac{du}{-2}=xdx\]\[\large -\frac{1}{2}\int\limits \frac{1}{\sqrt{u}}du\]\[\large -\frac{1}{2}\int\limits u^{-1/2}du = -\frac{1}{2}*2u^{1/2}= -\sqrt{1-x^2}+c\] put it all together. \[\large \int\limits\limits \frac{(2+x)}{\sqrt{1-x^2}}dx = -2\sin^{-1}(\sqrt{1-x^2})-\sqrt{1-x^2}+c\]
anonymous
  • anonymous
Why did you use that substitution for the first? It didn't help you at all...
anonymous
  • anonymous
Your answer is correct but it's not as aesthetically pleasing. Consider:|dw:1370830009490:dw| Here we know that \(\theta=\arcsin x\).
anonymous
  • anonymous
Now, also consider: |dw:1370830058934:dw|
anonymous
  • anonymous
Observe that $$\phi=\arcsin\sqrt{1-x^2}$$Remember that the sum of the interior angles of a triangle is \(\pi\):$$\theta+\phi+\frac\pi2=\pi\\\theta+\phi=\frac\pi2\\\phi=-\theta+\frac\pi2\\\arcsin\sqrt{1-x}^2=-\arcsin x+\frac\pi2\\-2\arcsin\sqrt{1-x^2}=2\arcsin x-\pi$$Now putting it all together, it is clear your solutions are equivalent up to a constant:$$-2\arcsin\sqrt{1-x^2}-\sqrt{1-x^2}=2\arcsin x-\sqrt{1-x}^2-\pi$$This constant \(-\pi\) is precisely the difference between your constants of integration.

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