int_{0}^{1} frac{d}{dx} left[ sin^{-1} left( | vedclass

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

Question

(C) Let $I = \int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$. By the Fundamental Theorem of Calculus,$\int_{a}^

Vedclass · Accepted Answer

$\pi/2$

Vedclass · Answer

(C) Let $I = \int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$. By the Fundamental Theorem of Calculus,$\int_{a}^{b} f'(x) dx = f(b) - f(a)$. Here,$f(x) = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$. So,$I = \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right]_{0}^{1}$. Evaluating at the limits: $I = \sin^{-1} \left( \frac{2(1)}{1+(1)^2} \right) - \sin^{-1} \left( \frac{2(0)}{1+(0)^2} \right)$. $I = \sin^{-1} \left( \frac{2}{2} \right) - \sin^{-1} (0)$. $I = \sin^{-1} (1) - \sin^{-1} (0)$. Since $\sin^{-1} (1) = \frac{\pi}{2}$ and $\sin^{-1} (0) = 0$,$I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

$\int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$ is equal to

Similar Questions

$\int_{\pi /4}^{3\pi /4} \frac{dx}{1 + \cos x}$ is equal to

$\int_0^{\frac{\pi}{4}} \sec^4 x \, dx =$

For $0 < x < \frac{\pi}{2}$,the integral $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \ln(e^{\cos x}) \, d(\sin x)$ is equal to:

$\int_2^3 \frac{\log x}{x} d x=$

$\int_0^1 \sin^{-1} x \, dx = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module