The value of the integral int_{-1}^{1} log_{e | vedclass

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(D) Let $I = \int_{-1}^{1} \log_{e}(\sqrt{1-x}+\sqrt{1+x}) dx$. Since the integrand $f(x) = \log_{e}(\sqrt{1-x}+\sqrt{1+x})$ is an even function,$I =

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$\log_{e} 2 + \frac{\pi}{2} - 1$

Vedclass · Answer

(D) Let $I = \int_{-1}^{1} \log_{e}(\sqrt{1-x}+\sqrt{1+x}) dx$. Since the integrand $f(x) = \log_{e}(\sqrt{1-x}+\sqrt{1+x})$ is an even function,$I = 2 \int_{0}^{1} \log_{e}(\sqrt{1-x}+\sqrt{1+x}) dx$.Using Integration by Parts $\int u dv = uv - \int v du$,let $u = \log_{e}(\sqrt{1-x}+\sqrt{1+x})$ and $dv = dx$.Then $du = \frac{1}{\sqrt{1-x}+\sqrt{1+x}} \cdot \left(\frac{-1}{2\sqrt{1-x}} + \frac{1}{2\sqrt{1+x}}\right) dx = \frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x^2}(\sqrt{1-x}+\sqrt{1+x})} dx = \frac{(\sqrt{1-x}-\sqrt{1+x})^2}{2\sqrt{1-x^2}(1-x-1-x)} dx = \frac{2-2\sqrt{1-x^2}}{-4x\sqrt{1-x^2}} dx = \frac{\sqrt{1-x^2}-1}{2x\sqrt{1-x^2}} dx$.$I = 2 \left[ x \log_{e}(\sqrt{1-x}+\sqrt{1+x}) \Big|_0^1 - \int_0^1 x \cdot \frac{\sqrt{1-x^2}-1}{2x\sqrt{1-x^2}} dx \right]$.$I = 2 \left[ (1 \cdot \log_{e}(\sqrt{2}) - 0) - \frac{1}{2} \int_0^1 \left(1 - \frac{1}{\sqrt{1-x^2}}\right) dx \right]$.$I = 2 \left[ \frac{1}{2} \log_{e} 2 - \frac{1}{2} (x - \sin^{-1} x) \Big|_0^1 \right]$.$I = \log_{e} 2 - (1 - \frac{\pi}{2}) = \log_{e} 2 + \frac{\pi}{2} - 1$.

The value of the integral $\int_{-1}^{1} \log_{e}(\sqrt{1-x}+\sqrt{1+x}) dx$ is equal to:

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