Question

Help!!!!
How do I integrate this:
integral (2x)/(sqrt1-x^2) from 0 to 1/2

Answer 1

\[\int\limits_{0}^{1/2}2x/\sqrt{1-x^2} dx\]

Answer 2

try \(u=1-x^2\)

Answer 3

\[\int\limits_{0}^{\frac{ 1 }{ 2 }}\frac{ 2x }{ \sqrt{1-x^2} }dx\]
put \[x^2=t,2 x dx=dt\]
when x=0,t=o
\[when~ x=\frac{ 1 }{ 2 },t=\frac{ 1 }{ 4 }\]
can you proceed further?

Answer 4

no I have no idea how to do this

Answer 5

you write the following:
"let \(u=1-x^2\) and so \(du=-2xdx\)

Answer 6

then take your integral and replace \(1-x^2\) by \(u\) and replace the \(2xdx\) by \(-du\)
pull the minus sign out front

Answer 7

that gives you
\[-\int \frac{du}{\sqrt{u}}\]

Answer 8

now find the anti derivative by writing this in exponential form and using the power rule backwards