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Question

01ddxsin-12x1+x2 dx is equal to

(a) 0
(b) π
(c) π/2
(d) π/4

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Solution

(c) π/2

We have,I=01ddxsin-12x1+x2dxWe know since f'(x) = f(x)f(x) =sin-12x1+x2 and f'(x)=ddxsin-12x1+x2 Therefore, I=sin-12x1+x201=sin-11-sin-10=π2

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