Integral of $\sqrt{1-x^2}$

Question

I don't understand why integral of $\sqrt{1-x^2}$ is : $1/2(x\times\sqrt{1+x^2}+\sinh^{-1}(x))$ and how can I calculate it with no help of computer ? tnx a lot

Answer 1

$x = \sin u, \quad dx = \cos u \; du$

$\displaystyle\int \sqrt{1 - x^2} \; dx$

$= \displaystyle\int\cos u \sqrt {1 - \sin ^2u} \; du$

$= \displaystyle\int\cos ^2 u \; du$

$= \displaystyle\int\dfrac{1}{2} + \dfrac {\cos 2 u}{2} \; du$

$= \dfrac{u}{2} + \dfrac{\sin 2u}{4}$

$= \dfrac{u}{2} + \dfrac{2\sin u \cos u}{4}$

$=\dfrac { \arcsin(x) }{2} + \dfrac {x\sqrt{1 - x^2}}{2} + c$

Answer 2

Hint: Multiply $\sqrt{1-x^2}$ by $\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}$

Answer 3

Another way: make the substitution $x=\cos(t)$.