Integral Calculus Question 100
Question: If $ \int_0^{1}{{{\cot }^{-1}}(1-x+x^{2})dx=\lambda \int_0^{1}{{{\tan }^{-1}}xdx,}} $ then $ \lambda $ is equal to
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \int_0^{1}{{{\cot }^{-1}}(1-x+x^{2})dx} $ $ =\int_0^{1}{{{\tan }^{-1}}( \frac{1}{1-x+x^{2}} )}dx $ $ =\int_0^{1}{{{\tan }^{-1}}( \frac{x+(1-x)}{1-x(1-x)} )}dx $ $ =\int_0^{1}{{{\tan }^{-1}}xdx+\int_0^{1}{{{\tan }^{-1}}(1-x)dx}} $ $ =\int_0^{1}{{{\tan }^{-1}}xdx+\int_0^{1}{{{\tan }^{-1}}[1-(1-x)]dx}} $ $ =2\int_0^{1}{{{\tan }^{-1}}xdx} $ or $ \lambda =2 $
BETA
