Question

${\int }_0^{1/2}{\cos }^{-1}xdx$ equal to

Answer 1

Consider the given integral.

$I={\int }_0^{1/2}{\cos }^{-1}xdx$

$I={\left(x\frac{-1}{\sqrt{1-x^2}}\right)}_0^{1/2}-{\int }_0^{1/2}\left(x\frac{-1}{\sqrt{1-x^2}}\right)dx$

$I=(\frac{-\frac{1}{2}}{\sqrt{1-{\left(\frac{1}{2}\right)}^2}}-0)+{\int }_0^{1/2}\left(\frac{x}{\sqrt{1-x^2}}\right)dx$

Let $t=1-x^2$

$\frac{dt}{dx}=0-2x$

$-\frac{dt}{2}=xdx$

Therefore,

$I=\left(\frac{-\frac{1}{2}}{\sqrt{1-{\left(\frac{1}{2}\right)}^2}}\right)-\frac{1}{2}{\int }_1^{3/4}\left(\frac{dt}{\sqrt{t}}\right)$

$I=\left(\frac{-\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}\right)-\frac{1}{2}{\left(2\sqrt{t}\right)}_1^{3/4}$

$I=\left(\frac{-\frac{1}{2}}{\sqrt{\frac{3}{4}}}\right)-(\sqrt{\frac{3}{4}}-1)$

$I=\left(\frac{-1}{\sqrt{3}}\right)-(\frac{\sqrt{3}}{2}-1)$

$I=\left(\frac{-1}{\sqrt{3}}\right)-\left(\frac{\sqrt{3}-2}{2}\right)$

$I=\frac{-2-3+2\sqrt{3}}{2\sqrt{3}}$

$I=\frac{-5+2\sqrt{3}}{2\sqrt{3}}$

Hence, this is the answer.