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Evaluate the following indefinite integral.

$$ \int \frac { 1 }{ \sqrt { 36-x^2 } } \, dx $$

How could i do this integral ?

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  • $\begingroup$ I assume you've learned trig substitution? What substitution would make that radical go away? $\endgroup$ – Mike Jan 9 '14 at 1:47
  • $\begingroup$ $ so\quad let\quad u\quad =\quad 36-{ x }^{ 2 }\quad \frac { du }{ dx } =-2x\quad du\quad =\quad -2xdx\quad dx=-\frac { du }{ 2x } \quad xdx\quad =\quad -\frac { 1 }{ 2 } \quad du $ Is that right so far ? $\endgroup$ – Out Of Bounds Jan 9 '14 at 1:50
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$$ \int\frac{1}{\sqrt{36-x^2}} \, dx = \int \frac{1}{\sqrt{36}{\sqrt{1-\dfrac{x^2}{36}}}} \, dx = \int \frac{1}{{\sqrt{1-\left(\dfrac{x}{6}\right)^2}}} \, \frac{dx}{6} = \int\frac{1}{\sqrt{1-u^2}} \, du. $$ That last integral is in all the tables.

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Substitute $x=6\sin\theta$, giving $dx=6\cos\theta\, d\theta$. So the integral changes to $$\int \frac{1}{\sqrt{36-36\sin^2\theta}}6\cos\theta \, d\theta$$

Now you should be able to do it

If you want to substitute $u=36-x^2$, then $du=-2x \, dx$. Then the integral becomes $\int\frac{1}{\sqrt{u}}\frac{du}{-2x}$. Again you will have to back substitute the value of $x$ in terms of $u$ from the original substitution which is tedious.

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  • $\begingroup$ I substituted differently. $ so\quad let\quad u\quad =\quad 36-{ x }^{ 2 }\quad \frac { du }{ dx } =-2x\quad du\quad =\quad -2xdx\quad dx=-\frac { du }{ 2x } \quad xdx\quad =\quad -\frac { 1 }{ 2 } \quad du $ Isn't that right ? $\endgroup$ – Out Of Bounds Jan 9 '14 at 1:55
  • $\begingroup$ no.check the answer $\endgroup$ – tattwamasi amrutam Jan 9 '14 at 2:04
  • $\begingroup$ Can you continue your answer with the trig substitution because i still can't continue with what you did. $\endgroup$ – Out Of Bounds Jan 9 '14 at 2:10
  • $\begingroup$ @Tennisman note that $1 - \sin^2 \theta = \cos^2 \theta$ $\endgroup$ – DanZimm Jan 9 '14 at 2:52

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