If I=∫10cot−1(1−x+x2)dx=k∫10tan−1xdx, then the value of k equals
A
1
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B
2
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C
π
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D
2π
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Solution
The correct option is B 2 We have, I=∫10cot−1(1−x+x2)dx=k∫10tan−1xdx⇒∫10tan−1(1(1−x+x2)}dx=k∫10tan−1xdx⇒∫10tan−1(11−x(1−x)}dx=k∫10tan−1xdx⇒∫10tan−1((1−x)+x1−x(1−x)}dx=k∫10tan−1xdx⇒∫10(tan−1(1−x)+tan−1x}dx=k∫10tan−1xdx⇒∫10tan−1(1−x)dx+∫10tan−1xdx=k∫10tan−1xdx⇒∫10tan−1(1−(1−x)]dx+∫10tan−1xdx=k∫10tan−1xdx⇒∫10tan−1xdx+∫10tan−1xdx=k∫10tan−1xdx⇒2∫10tan−1xdx=k∫10tan−1xdx⇒k=2