Question

Little Integral Confusion, Posting in a minute

Answer 1

\[\large \int\limits_{}^{} \frac{5}{\sqrt{4 - 9x^2}}dx\]
>Factor out a 5...
\[\large 5\int\limits_{}^{} \frac{1}{\sqrt{4 - 9x^2}}dx\]
>Factor a 4 from the bottom
\[\large 5\int\limits_{}^{} \frac{1}{\sqrt{4(1 - \frac{9}{4}x)^2}}dx\]
>Simplify
\[\large \frac{5}{2}\int\limits_{}^{} \frac{1}{\sqrt{1 - \frac{9}{4}x^2}}dx\]
>Let u = 3/2x
>so du = 3/2dx
>dx = 2/3du
\[\large \frac{5}{2}\int\limits_{}^{} \frac{1}{\sqrt{1 - u^2}}\frac{2}{3}du\]
>Factor out the 2/3
\[\large \frac{10}{9}\int\limits_{}^{} \frac{1}{\sqrt{1 - u^2}}du\]
>>>\[\int\limits_{}^{} \frac{1}{\sqrt{1 - u^2}} = \sin^{-1} (u)\]<<< so
\[\large \frac{10}{9}\sin^{-1}(u) + C \]
>u = 3/2x

\[\large \frac{10}{9}\sin^{-1}(\frac{3x}{2}) + C \]

This is what I get for an answer but according to my text it is supposed to be

\[\large \frac{5}{3}\sin^{-1}(\frac{3x}{2}) + C \]

Where am I messing up?

Answer 2

your 3rd line looks odd

Answer 3

\[\large 5\int\limits_{}^{} \frac{1}{\sqrt{4 - 9x^2}}dx\]
\[\large 5\int\limits_{}^{} \frac{1}{\sqrt{4(1 - \frac{9}{4}x^2)}}dx\]
\[\large 5\int\limits_{}^{} \frac{1}{\sqrt{4(1 - (\frac{3}{2}x)^2)}}dx\]

Answer 4

i think you made up for it ... might of been a typo

let sin(u) = 3/2 x might be a nice subsitution

Answer 5

to me every thing is correct just how 5/2*2/3= 10/9??? it's 5/3

Answer 6

this is the mistake ^

there is no mistake :P

Answer 7

SUCH A SIMPLE MISTAKE!!!! -_-

Answer 8

\[ 5\int\limits_{}^{} \frac{1}{\sqrt{4(1 - (\frac{3}{2}x)^2)}}dx\]
\[ \frac52\int\limits_{}^{} \frac{1}{\sqrt{1 - (\frac{3}{2}x)^2}}dx\]
\[ \frac52\int\limits_{}^{} \frac{1}{\sqrt{1 - sin^2(u)}}cos(u)\frac23du\]
\[ \frac52\int\limits_{}^{} \frac{cos(u)}{cos(u)}\frac23du\]

Answer 9

Thank you @Loser66 and @amistre64 I see where you're going with that trig sub...another great way to look at it! Thanks guys!

Answer 10

good luck ;)