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Q.
If $\int\limits_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right) d x=\lambda \int\limits_{0}^{1} \tan ^{-1} x d x$ then ' $\lambda$ ' is equal to
Integrals
Solution:
$\int\limits_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right) d x$
$=\int\limits_{0}^{1} \tan ^{-1}\left(\frac{1}{1-x+x^{2}}\right) d x=\int\limits_{0}^{1} \tan ^{-1}\left(\frac{x+(1-x)}{1-x(1-x)}\right) d x$
$=\int\limits_{0}^{1} \tan ^{-1} x d x+\int\limits_{0}^{1} \tan ^{-1}(1-x) d x$
$=\int\limits_{0}^{1} \tan ^{-1} x d x+\int\limits_{0}^{1} \tan ^{-1}[1-(1-x)] d x=2 \int\limits_{0}^{1} \tan ^{-1} x d x$
$\Rrightarrow \lambda=2$