Question

\[\int_{0}^{1} \sqrt(1+\frac{x^2}{4-x^2})\]

Answer 1

Is it??

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

Answer 2

sorry no it's 4-x^2

Answer 3

So is it?

\[\int\limits_{0}^{1} \sqrt{1+\frac{x^2}{4-x^2}}\]

Answer 4

Take the LCM of the denominator...

\[\frac{4 - x^2 + x^2}{4 - x^2} = \frac{4}{4 - x^2}\]

Answer 5

doesn't this become

\[\int\limits_{0}^{1} \sqrt{\frac{4}{4 - x^2}} dx\]

Answer 6

which can be written as

\[2\int\limits_{0}^{1}\sqrt{\frac{1}{4 -x^2}} dx\]

Answer 7

and \[2\int\limits_{0}^{1}\frac{1}{\sqrt{4 -x^2}} dx\]

Answer 8

Use the following formula:

\[\int\limits_{}^{}\frac{1}{\sqrt{a^2 - x^2}}.dx = \sin^{-1}\frac{x}{a} + C\]

Here a = 2..

Answer 9

got it, thanks!

Answer 10

Welcome dear..