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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 =

Vedclass · Accepted Answer

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

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:

Similar Questions

If $f(x) = \frac{x^3+5}{\sqrt{12+x}}$ and $\int_{-5}^5 f(x) dx = \int_0^5 (f(x) + g(x)) dx$,then $g(x) =$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$.

$\int_{-1}^1 \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} dx$ is equal to

$\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\cot x}} = $ . . . . . . .

$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module