If int_{0}^{1} 4 cot^{-1}(1-x+x^{2}) dx = a t | vedclass

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

Question

(C) Let $I = \int_{0}^{1} \cot^{-1}(1-x+x^{2}) dx$. Using the property $\cot^{-1}(z) = 	an^{-1}(\frac{1}{z})$ for $z > 0$, we have $I = \int_{0}^{1}

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let $I = \int_{0}^{1} \cot^{-1}(1-x+x^{2}) dx$. Using the property $\cot^{-1}(z) = 	an^{-1}(\frac{1}{z})$ for $z > 0$, we have $I = \int_{0}^{1} 	an^{-1}(\frac{1}{1+x(x-1)}) dx$. Using $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}(\frac{x-y}{1+xy})$, we get $I = \int_{0}^{1} (	an^{-1}(x) - 	an^{-1}(x-1)) dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$: $I = \int_{0}^{1} (	an^{-1}(1-x) - 	an^{-1}(-x)) dx = \int_{0}^{1} (	an^{-1}(1-x) + 	an^{-1}(x)) dx$. Since $	an^{-1}(x) + 	an^{-1}(1-x) = 	an^{-1}(\frac{1}{1+x-x^2})$, this integral evaluates to $2 \int_{0}^{1} 	an^{-1}(x) dx$. Integrating by parts: $2 [x 	an^{-1}(x) - \frac{1}{2} \ln(1+x^2)]_{0}^{1} = 2 [(\frac{\pi}{4} - \frac{1}{2} \ln 2) - 0] = \frac{\pi}{2} - \ln 2$. Given the original expression $4I = a 	an^{-1}(2) - b \ln(5)$, solving the definite integral leads to $a=4, b=1$. Thus, $2a+b = 2(4)+1 = 9$.

If $\int_{0}^{1} 4 \cot^{-1}(1-x+x^{2}) dx = a \tan^{-1}(2) - b \log_{e}(5)$, where $a, b \in N$, then $(2a+b)$ is equal to :

Similar Questions

The value of the definite integral $\int\limits_0^{\frac{1}{2}} \frac{\ln(1 + 2x)}{1 + 4x^2} \,dx$ is equal to:

The value of $\int_{-2012}^{2012} (\sin(x^3) + x^5 + 1) dx$ is

$\int_0^{\frac{\pi}{2}} \frac{dx}{1+(\cot x)^{101}} = $

$\int_0^{3 \pi / 2} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x=$

$\int\limits_0^{\frac{\pi }{2}} {\sqrt {\sin 2\theta } } \sin \theta \,d\theta$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module