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Mathematics Question on Definite Integral

The value of integral ∫01 1−x1+xdx\int\limits_0^1 \, \sqrt{\frac{1-x}{1+x}}dx is

A

π2+1\frac{\pi}{2} + 1

B

π2−1\frac{\pi}{2} - 1

C

−1-1

D

11

Answer

π2−1\frac{\pi}{2} - 1

Explanation

Solution

Let I=∫011−x1+xdxI = \int\limits^{1}_{0} \sqrt{\frac{1-x}{1+x}} dx
=∫011−x1−x2dx= \int\limits^{1}_{0} \frac{1-x}{\sqrt{1-x^{2}}}dx
=∫0111−x2dx−∫01x1+x2dx= \int\limits^{1}_{0} \frac{1}{\sqrt{1-x^{2}}} dx - \int\limits^{1}_{0} \frac{x}{\sqrt{1+x^{2}}}dx
=[sin⁡−1x]01−∫01x1−x2dx= \left[\sin^{-1}x\right]^{1}_{0} - \int\limits^{1}_{0} \frac{x}{\sqrt{1-x^{2}}} dx
Put t2=1−x2t^{2} = 1 -x^{2}
⇒ 2t dt=−2x dx\Rightarrow \, 2t \, dt = - 2x \, dx
⇒ t dt=−x dx\Rightarrow \, t \, dt = -x \, dx
∴I=(sin⁡−11−sin⁡−10)+∫10ttdt\therefore I = \left(\sin^{-1} 1 - \sin^{-1}0\right)+\int\limits_{1}^{0} \frac{t}{t} dt
=π2+[t]10=π2−1= \frac{\pi}{2} + \left[t\right]^{0}_{1} = \frac{\pi}{2} - 1
Let I=∫011−x1+xdxI=\int\limits_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x
Put x=cos⁡2 θx=\cos 2 \,\theta
⇒dx=−2sin⁡2 θ d θ\Rightarrow d x=-2 \sin 2 \,\theta \,d \,\theta
∴I=−∫π/401−cos⁡2θ1+cos⁡2θ⋅2sin⁡θ d θ\therefore I=-\int\limits_{\pi / 4}^{0} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}} \cdot 2 \sin \theta \,d\, \theta
=−2∫π/40sin⁡θcos⁡θ⋅2sin⁡θcos⁡θ d θ=-2 \int\limits_{\pi / 4}^{0} \frac{\sin \theta}{\cos \theta} \cdot 2 \sin \theta \cos \theta \,d \,\theta
=−2∫π/40(1−cos⁡2θ) d θ=-2 \int\limits_{\pi / 4}^{0}(1-\cos 2 \theta) \,d \,\theta
=−2[θ−sin⁡2θ2]π/40=-2\left[\theta-\frac{\sin 2 \theta}{2}\right]_{\pi / 4}^{0}
=−2[0−(π4−12)]=-2\left[0-\left(\frac{\pi}{4}-\frac{1}{2}\right)\right]
=π2−1=\frac{\pi}{2}-1