We need
∫ 1 x 1 − x 1 + x d x = g ( x ) + C , \int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,dx = g(x)+C, ∫ x 1 1 + x 1 − x d x = g ( x ) + C , with g ( 1 ) = 0 g(1)=0 g ( 1 ) = 0 g ( 1 2 ) g\left(\frac12\right) g ( 2 1 )
Use the standard substitution
x = cos θ . x=\cos\theta. x = cos θ . d x = − sin θ d θ , dx=-\sin\theta\,d\theta, d x = − sin θ d θ , 1 − x 1 + x = 1 − cos θ 1 + cos θ = tan θ 2 . \sqrt{\frac{1-x}{1+x}}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\tan\frac\theta2. 1 + x 1 − x = 1 + c o s θ 1 − c o s θ = tan 2 θ . x = cos θ . x=\cos\theta. x = cos θ .
1 cos θ tan θ 2 ( − sin θ ) d θ . \frac{1}{\cos\theta}\tan\frac\theta2(-\sin\theta)\,d\theta. cos θ 1 tan 2 θ ( − sin θ ) d θ . Now use
tan θ 2 = sin θ 1 + cos θ . \tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta}. tan 2 θ = 1 + c o s θ s i n θ .
1 cos θ ⋅ sin θ 1 + cos θ ⋅ ( − sin θ ) = − sin 2 θ cos θ ( 1 + cos θ ) . \frac{1}{\cos\theta}\cdot \frac{\sin\theta}{1+\cos\theta}\cdot (-\sin\theta)
= -\frac{\sin^2\theta}{\cos\theta(1+\cos\theta)}. cos θ 1 ⋅ 1 + cos θ sin θ ⋅ ( − sin θ ) = − cos θ ( 1 + cos θ ) sin 2 θ . Since
sin 2 θ = ( 1 − cos θ ) ( 1 + cos θ ) , \sin^2\theta=(1-\cos\theta)(1+\cos\theta), sin 2 θ = ( 1 − cos θ ) ( 1 + cos θ ) ,
− 1 − cos θ cos θ = 1 − sec θ . -\frac{1-\cos\theta}{\cos\theta} = 1-\sec\theta. − cos θ 1 − cos θ = 1 − sec θ . Therefore,
∫ 1 x 1 − x 1 + x d x = ∫ ( 1 − sec θ ) d θ = θ − ∫ sec θ d θ . \int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,dx
=\int (1-\sec\theta)\,d\theta
=\theta-\int \sec\theta\,d\theta. ∫ x 1 1 + x 1 − x d x = ∫ ( 1 − sec θ ) d θ = θ − ∫ sec θ d θ . And
∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ . \int \sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|. ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣. So
g ( x ) = θ − ln ∣ sec θ + tan θ ∣ + C , g(x)=\theta-\ln|\sec\theta+\tan\theta|+C, g ( x ) = θ − ln ∣ sec θ + tan θ ∣ + C , where x = cos θ x=\cos\theta x = cos θ
Express everything in terms of x x x
Since x = cos θ x=\cos\theta x = cos θ
θ = cos − 1 x , \theta=\cos^{-1}x, θ = cos − 1 x , sec θ = 1 x , \sec\theta=\frac1x, sec θ = x 1 , tan θ = 1 − x 2 x . \tan\theta=\frac{\sqrt{1-x^2}}{x}. tan θ = x 1 − x 2 . Thus
sec θ + tan θ = 1 + 1 − x 2 x . \sec\theta+\tan\theta=\frac{1+\sqrt{1-x^2}}{x}. sec θ + tan θ = x 1 + 1 − x 2 . Hence
g ( x ) = cos − 1 x − ln ( 1 + 1 − x 2 x ) + C . g(x)=\cos^{-1}x-\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)+C. g ( x ) = cos − 1 x − ln ( x 1 + 1 − x 2 ) + C .
Use the condition g ( 1 ) = 0 g(1)=0 g ( 1 ) = 0
At x = 1 x=1 x = 1
cos − 1 ( 1 ) = 0 , 1 − 1 2 = 0 , ln ( 1 + 0 1 ) = ln 1 = 0. \cos^{-1}(1)=0,
\qquad \sqrt{1-1^2}=0,
\qquad \ln\left(\frac{1+0}{1}\right)=\ln 1=0. cos − 1 ( 1 ) = 0 , 1 − 1 2 = 0 , ln ( 1 1 + 0 ) = ln 1 = 0. So C = 0 C=0 C = 0
Therefore,
g ( x ) = cos − 1 x − ln ( 1 + 1 − x 2 x ) . g(x)=\cos^{-1}x-\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right). g ( x ) = cos − 1 x − ln ( x 1 + 1 − x 2 ) .
Now evaluate at x = 1 2 x=\frac12 x = 2 1
First,
cos − 1 ( 1 2 ) = π 3 . \cos^{-1}\left(\frac12\right)=\frac\pi3. cos − 1 ( 2 1 ) = 3 π . Also,
1 − ( 1 2 ) 2 = 3 4 = 3 2 . \sqrt{1-\left(\frac12\right)^2}=\sqrt{\frac34}=\frac{\sqrt3}{2}. 1 − ( 2 1 ) 2 = 4 3 = 2 3 . So
1 + 1 − x 2 x = 1 + 3 2 1 2 = 2 + 3 . \frac{1+\sqrt{1-x^2}}{x}
=\frac{1+\frac{\sqrt3}{2}}{\frac12}
=2+\sqrt3. x 1 + 1 − x 2 = 2 1 1 + 2 3 = 2 + 3 . Thus
g ( 1 2 ) = π 3 − ln ( 2 + 3 ) . g\left(\frac12\right)=\frac\pi3-\ln(2+\sqrt3). g ( 2 1 ) = 3 π − ln ( 2 + 3 ) . Now rationalize:
2 + 3 = 3 + 1 3 − 1 , 2+\sqrt3=\frac{\sqrt3+1}{\sqrt3-1}, 2 + 3 = 3 − 1 3 + 1 , because
3 + 1 3 − 1 = ( 3 + 1 ) 2 3 − 1 = 4 + 2 3 2 = 2 + 3 . \frac{\sqrt3+1}{\sqrt3-1}
=\frac{(\sqrt3+1)^2}{3-1}
=\frac{4+2\sqrt3}{2}=2+\sqrt3. 3 − 1 3 + 1 = 3 − 1 ( 3 + 1 ) 2 = 2 4 + 2 3 = 2 + 3 . Hence
− ln ( 2 + 3 ) = ln ( 3 − 1 3 + 1 ) . -\ln(2+\sqrt3)=\ln\left(\frac{\sqrt3-1}{\sqrt3+1}\right). − ln ( 2 + 3 ) = ln ( 3 + 1 3 − 1 ) . Therefore,
g ( 1 2 ) = ln ( 3 − 1 3 + 1 ) + π 3 . g\left(\frac12\right)=\ln\left(\frac{\sqrt3-1}{\sqrt3+1}\right)+\frac\pi3. g ( 2 1 ) = ln ( 3 + 1 3 − 1 ) + 3 π .
Compare with options:
This matches Option A .