Evaluate the integral 3/sqrt (4-9x^2)dx, I got to the point of getting u=x du=1dx and a=sqrt(4/9) and i used the trig fcn of arcsinu/a+c but i still can't figure it out. please help!

Question

turingtest · Answer

$$\int{3\over\sqrt{4-9x^2}}dx$$is the prob?

anonymous · Answer

yeah

turingtest · Answer

then you need a trigonometric substitution$$\frac23x=\sin\theta$$do you know how to do these from here?

mimi_x3 · Answer

$$\int\limits\frac{3}{\sqrt{4-9x^{2}}}  dx =>\int\limits\frac{3}{\sqrt{9(\frac{4}{9})-x^{2})}}  dx$$
i think that i answered this question before..

turingtest · Answer

whoops, bad sub, let me fix that...

mimi_x3 · Answer

No subsitution is needed i think..

turingtest · Answer

the sub you need is$$x=\frac23\sin\theta$$@Mimi_x3 please demonstrate

anonymous · Answer

could you show me the steps? because my answer was 3arcsin(9xsqrt(4/9)/4 +c but i don't know if that is right?

turingtest · Answer

close, should be arcsin(3x/2)+C

mimi_x3 · Answer

$$\frac{3}{9}  \int\limits\frac{1}{\sqrt{(\frac{4}{9})-x^{2}}}  dx =>\frac{1}{3}  \int\limits\frac{1}{\frac{2^{2}}{3^{2}}-x^{2}}  dx $$
this is the step..then it's a trig inverse function..

anonymous · Answer

@mimi_x3 how did you pull out the 9 out of sq rt to get the coefficient of 3/9?

turingtest · Answer

$$\sqrt{4-9x^2}=\sqrt{9(\frac49-x^2})=3\sqrt{\frac49-x^2}$$

mimi_x3 · Answer

lol, i think that something wrong..which i can't figure out at this time.

turingtest · Answer

and nice job Mimi, never would have thought to do it that way...
you dropped the sqrt sign at the end though, but you still did the problem right

mimi_x3 · Answer

ohh okay, small mistake my bad..