If ∫01 cot−1⁡1−x+x2dx=k∫01 tan−1⁡xdx, then k=

Let I=∫01 cot−1⁡1−x+x2dxI=∫01 tan−1⁡11−x(1−x)dx=∫01 tan−1⁡x+(1−x)1−x(1−x)dxI=∫01 tan−1⁡x+tan−1⁡(1−x)dxI=∫01 tan−1⁡xdx+∫01 tan−1⁡(1−x)dxI=∫01 tan−1⁡xdx+∫01 tan−1⁡(1−(1−x))dx=2∫01 tan−1⁡xdx∴k=2

If ∫01 cot−1⁡1−x+x2dx=k∫01 tan−1⁡xdx, then k=

Let I=∫01 cot−1⁡1−x+x2dxI=∫01 tan−1⁡11−x(1−x)dx=∫01 tan−1⁡x+(1−x)1−x(1−x)dxI=∫01 tan−1⁡x+tan−1⁡(1−x)dxI=∫01 tan−1⁡xdx+∫01 tan−1⁡(1−x)dxI=∫01 tan−1⁡xdx+∫01 tan−1⁡(1−(1−x))dx=2∫01 tan−1⁡xdx∴k=2

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If $\int_{0}^{1} \cot^{- 1} (1 - x + x^{2}) d x = k \int_{0}^{1} \tan^{- 1} x d x,$ then $k =$

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$π$

$2 π$

answer is B.

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Detailed Solution

Let $I = \int_{0}^{1} \cot^{- 1} (1 - x + x^{2}) d x$

$\begin{array}{l} I = \int_{0}^{1} \tan^{- 1} \{\frac{1}{1 - x (1 - x)}\} d x = \int_{0}^{1} \tan^{- 1} \{\frac{x + (1 - x)}{1 - x (1 - x)}\} d x \\ I = \int_{0}^{1} \{\tan^{- 1} x + \tan^{- 1} (1 - x)\} d x \\ I = \int_{0}^{1} \tan^{- 1} x d x + \int_{0}^{1} \tan^{- 1} (1 - x) d x \\ I = \int_{0}^{1} \tan^{- 1} x d x + \int_{0}^{1} \tan^{- 1} (1 - (1 - x)) d x = 2 \int_{0}^{1} \tan^{- 1} x d x \\ ∴ k = 2 \end{array}$