Question

If $I={\int }_0^{1}cos\left(2Co{t}^{-1}\sqrt{\frac{1-x}{1+x}}\right)dxthen$

• A. $I>\frac{1}{2}$
• B. $I=-\frac{1}{2}$
• C. $I<\frac{1}{2}$
• D. $I=\frac{1}{2}$

Answer 1

The correct option is B $I=-\frac{1}{2}$

$I={\int }_0^{1}cos\left(2co{t}^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$
$Putx=cos\theta \Rightarrow dx=-sin\theta d\theta ;x=0,1\Rightarrow \theta =\frac{\pi }{2},0$
$I=-{\int }_{\frac{\pi }{2}}^{0}cos\left(2co{t}^{-1}\sqrt{\frac{2si{n}^2\frac{\theta }{2}}{2co{s}^2\frac{\theta }{2}}}\right)sin\theta d\theta ={\int }_0^{\frac{\pi }{2}}cos\left[2co{t}^{-1}\left(tan\frac{\theta }{2}\right)\right]sin\theta d\theta$
$={\int }_{\frac{\pi }{2}}^{0}cos\left(2co{t}^{-1}cot(\frac{\pi }{2}-\frac{\theta }{2})\right)sin\theta d\theta ={\int }_0^{\frac{\pi }{2}}cos\left[2(\frac{\pi }{2}-\frac{\theta }{2})\right]sin\theta d\theta$
$={\int }_0^{\frac{\pi }{2}}cos(\pi -\theta )sin\theta d\theta$
$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}2sin\theta cos\theta d\theta =-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}sin2\theta d\theta =\frac{1}{4}[cos2{\theta }_0^{\frac{\pi }{2}}=\frac{1}{4}(-1-1)=-\frac{1}{2}$