# Indefinite Integral Questions

Evaluate the following indefinite integral.

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

How could i do this integral ?

• I assume you've learned trig substitution? What substitution would make that radical go away? – Mike Jan 9 '14 at 1:47
• $so\quad let\quad u\quad =\quad 36-{ x }^{ 2 }\quad \frac { du }{ dx } =-2x\quad du\quad =\quad -2xdx\quad dx=-\frac { du }{ 2x } \quad xdx\quad =\quad -\frac { 1 }{ 2 } \quad du$ Is that right so far ? – Out Of Bounds Jan 9 '14 at 1:50

$$\int\frac{1}{\sqrt{36-x^2}} \, dx = \int \frac{1}{\sqrt{36}{\sqrt{1-\dfrac{x^2}{36}}}} \, dx = \int \frac{1}{{\sqrt{1-\left(\dfrac{x}{6}\right)^2}}} \, \frac{dx}{6} = \int\frac{1}{\sqrt{1-u^2}} \, du.$$ That last integral is in all the tables.
Substitute $x=6\sin\theta$, giving $dx=6\cos\theta\, d\theta$. So the integral changes to $$\int \frac{1}{\sqrt{36-36\sin^2\theta}}6\cos\theta \, d\theta$$
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.
• 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 ? – Out Of Bounds Jan 9 '14 at 1:55
• @Tennisman note that $1 - \sin^2 \theta = \cos^2 \theta$ – DanZimm Jan 9 '14 at 2:52