JEE Main 2022 — Mathematics Calculus

Given,

$$\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}dx=g(x)+c$$∫x1​1+x1−x​​dx=g(x)+c

Put $x=\mathrm{cos}2\theta$x=cos2θ

$$dx=-2\mathrm{sin}2\theta \cdot d\theta$$dx=−2sin2θ⋅dθ

$$=\int \frac{1}{\mathrm{cos}2\theta }\mathrm{tan}\theta (-4\mathrm{sin}\theta \cdot \mathrm{cos}\theta )d\theta$$=∫cos2θ1​tanθ(−4sinθ⋅cosθ)dθ

$$=\int \frac{1}{\mathrm{cos}2\theta }(-4{\mathrm{sin}}^{2}\theta )d\theta$$=∫cos2θ1​(−4sin2θ)dθ

$$=-2\int \frac{1-\mathrm{cos}2\theta }{\mathrm{cos}2\theta }d\theta$$=−2∫cos2θ1−cos2θ​dθ

$$=-\frac{2}{2}\mathrm{ln}|\mathrm{sec}2\theta -\mathrm{tan}2\theta |+2\theta +c$$=−22​ln∣sec2θ−tan2θ∣+2θ+c

$$=\mathrm{ln}|\mathrm{sec}2\theta -\mathrm{tan}2\theta |+2\theta +c$$=ln∣sec2θ−tan2θ∣+2θ+c

$$=\mathrm{ln}|\frac{1-\mathrm{sin}2\theta }{\mathrm{cos}2\theta }|+{\mathrm{cos}}^{-1}x+c$$=ln∣cos2θ1−sin2θ​∣+cos−1x+c

$$\underset{g(x)}{\underset{⏟}{=\mathrm{ln}|\frac{1-\sqrt{1-{x}^{2}}}{x}|+{\mathrm{cos}}^{-1}x+c}}$$g(x)⏟=ln∣x1−1−x2​​∣+cos−1x+c​​

$$\therefore g(1)=0\Rightarrow c=0$$∴g(1)=0⇒c=0

So, $g(x)=\mathrm{ln}|\frac{1-\sqrt{1-{x}^{2}}}{x}|+{\mathrm{cos}}^{-1}x$g(x)=ln∣x1−1−x2​​∣+cos−1x

$$g(\frac{1}{2})=\mathrm{ln}|2-\sqrt{3}|+\frac{\pi }{3}$$g(21​)=ln∣2−3​∣+3π​

$$g(\frac{1}{2})=\mathrm{ln}|\frac{\sqrt{3}-1}{\sqrt{3}+1}|+\frac{\pi }{3}$$g(21​)=ln∣3​+13​−1​∣+3π​