Question

another integration question. integrate (9x^2)/(sqrt (1-x^2)) dx

Answer 1

I think you need trig sub

Answer 2

i've try but don't get it :(

Answer 3

try substitute x=sin u and x=cos u du

Then \[\sqrt{1-x^2}=\sqrt{1-\sin ^2u }=\cos u\]
\[u=\sin ^{-1}(x)\]

\[9\int\limits \sin ^2(u) \, du\]
Use half angle formula
\[9\int\limits \left(\frac{1}{2}-\frac{1}{2} \cos (2 u)\right) \, du\]
\[\text{}=9\int\limits \frac{1}{2} \, du-\frac{9}{2}\int\limits \cos (2 u) \, du\]
for cos2u, use v=2u dv=2du
\[9\int\limits \frac{1}{2} \, du-\frac{9}{4}\int\limits \cos v \, dv\]
integrate separately,
\[\frac{9 u}{2}-\frac{9 \sin (s)}{4}+c\]
sub back those leftovers
\[\frac{9}{2} \sin ^{-1}(x)-\frac{9}{2} x \sqrt{1-x^2}\]

Answer 4

if we don't use trigo can get the answer or not?

Answer 5

i don't think so

Answer 6

but the answer is -6(1-x^2)^ (1/2) +c

Answer 7

you sure you wrote the question correctly? if yes then im not sure, but this is my work

Answer 8

\[\int\limits_{}^{} (9x^2)/\sqrt{1-x^2} dx\]