JEE Main 2021 — Mathematics Calculus

Let$\displaystyle I={\int }_{-1}^{1}{\mathrm{log}}_{e}(\sqrt{1-x}+\sqrt{1+x})dx$I=∫−11​loge​(1−x​+1+x​)dx

$$\Rightarrow I=2{\int }_{0}^{1}\underset{I}{\underset{⏟}{{\mathrm{log}}_{e}(\sqrt{1-x}+\sqrt{1+x})}}\cdot \underset{II}{\underset{⏟}{1}}dx$$⇒I=2∫01​I⏟loge​(1−x​+1+x​)​​⋅II⏟1​​dx

$$\Rightarrow I=2[{(x{\mathrm{log}}_{e}(\sqrt{1-x}+\sqrt{1+x}))}_{0}^{1}-{\int }_{0}^{1}x\cdot (\frac{1}{\sqrt{1-x}+\sqrt{1+x}})\cdot (\frac{1}{2\sqrt{1+x}}-\frac{1}{2\sqrt{1-x}})dx]$$⇒I=2[(xloge​(1−x​+1+x​))01​−∫01​x⋅(1−x​+1+x​1​)⋅(21+x​1​−21−x​1​)dx]

$$\Rightarrow I=2{\mathrm{log}}_{e}\sqrt{2}-\frac{2}{2}{\int }_{0}^{1}{\frac{x(\sqrt{1-x}-\sqrt{1+x})}{(\sqrt{1-x}+\sqrt{1+x})\sqrt{1-{x}^{2}}}}dx$$⇒I=2loge​2​−22​∫01​(1−x​+1+x​)1−x2​x(1−x​−1+x​)​dx

$$\Rightarrow I=2{\mathrm{log}}_{e}\sqrt{2}-{\int }_{0}^{1}{\frac{x{(\sqrt{1-x}-\sqrt{1+x})}^{2}}{-2x\sqrt{1-{x}^{2}}}}dx$$⇒I=2loge​2​−∫01​−2x1−x2​x(1−x​−1+x​)2​dx

$$\Rightarrow I=({\mathrm{log}}_{e}2)+\frac{1}{2}{\int }_{0}^{1}\frac{(2-2\sqrt{1-{x}^{2}})}{\sqrt{1-{x}^{2}}}dx$$⇒I=(loge​2)+21​∫01​1−x2​(2−21−x2​)​dx

$$\Rightarrow I=({\mathrm{log}}_{e}2)+{\int }_{0}^{1}(\frac{1-\sqrt{1-{x}^{2}}}{\sqrt{1-{x}^{2}}})dx$$⇒I=(loge​2)+∫01​(1−x2​1−1−x2​​)dx

$$\Rightarrow I=({\mathrm{log}}_{e}2)+{({\mathrm{sin}}^{-1}x-x)}_{0}^{1}$$⇒I=(loge​2)+(sin−1x−x)01​

$$\Rightarrow I=({\mathrm{log}}_{e}2)+\frac{\pi }{2}-1$$⇒I=(loge​2)+2π​−1