If 2 int_0^1 tan^{-1} x , dx = int_0^1 cot^{- | vedclass

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

Question

(B) We are given the identity $2 \int_{0}^{1} 	an^{-1} x \, dx = \int_{0}^{1} \cot^{-1} (1 - x + x^2) \, dx$. Using the property $\cot^{-1} u = \frac

Vedclass · Accepted Answer

$\log 2$

Vedclass · Answer

(B) We are given the identity $2 \int_{0}^{1} 	an^{-1} x \, dx = \int_{0}^{1} \cot^{-1} (1 - x + x^2) \, dx$. Using the property $\cot^{-1} u = \frac{\pi}{2} - 	an^{-1} u$,we can rewrite the right side: $2 \int_{0}^{1} 	an^{-1} x \, dx = \int_{0}^{1} \left( \frac{\pi}{2} - 	an^{-1} (1 - x + x^2) ight) dx$. $2 \int_{0}^{1} 	an^{-1} x \, dx = \frac{\pi}{2} - \int_{0}^{1} 	an^{-1} (1 - x + x^2) \, dx$. Rearranging the terms,we get: $\int_{0}^{1} 	an^{-1} (1 - x + x^2) \, dx = \frac{\pi}{2} - 2 \int_{0}^{1} 	an^{-1} x \, dx$. Now,we evaluate $I = \int_{0}^{1} 	an^{-1} x \, dx$ using integration by parts: $I = [x 	an^{-1} x]_0^1 - \int_0^1 \frac{x}{1+x^2} \, dx = \frac{\pi}{4} - \frac{1}{2} [\ln(1+x^2)]_0^1 = \frac{\pi}{4} - \frac{1}{2} \ln 2$. Substituting this back into our expression: $\int_{0}^{1} 	an^{-1} (1 - x + x^2) \, dx = \frac{\pi}{2} - 2 \left( \frac{\pi}{4} - \frac{1}{2} \ln 2 ight) = \frac{\pi}{2} - \frac{\pi}{2} + \ln 2 = \ln 2$.

If $2 \int_0^1 \tan^{-1} x \, dx = \int_0^1 \cot^{-1} (1 - x + x^2) \, dx,$ then $\int_0^1 \tan^{-1} (1 - x + x^2) \, dx$ is equal to

Similar Questions

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

The value of $\int_0^{\pi / 2} \frac{d x}{1+(\tan x)^{\sqrt{2018}}}$ is equal to

$\int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}\,dx = } $

If $f : R \rightarrow R$ is a continuous function satisfying $\int \limits_0^{\pi / 2} f(\sin 2x) \cdot \sin x \, dx + \alpha \int \limits_0^{\pi / 4} f(\cos 2x) \cdot \cos x \, dx = 0$,then $\alpha$ is equal to

$I = \int\limits_0^1 \sqrt[3]{2x^3 - 3x^2 - x + 1} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module