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

If $f$ and $g$ are continuous functions on $[0, a]$ satisfying $f(x) = f(a - x)$ and $g(x) + g(a - x) = 2$,then $\int_0^a f(x)g(x) dx = $

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{3} + x \cos x + \tan^{5} x + 1) dx$ is

$\int_0^\pi x \sin x \, dx = $

If $n$ is any integer,then $\int_0^\pi {e^{\cos^2 x} \cos^3((2n + 1)x)} \, dx = $

The value of $\int_0^{\pi /2} \frac{dx}{1 + \tan^3 x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module