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Question

π/2011+4tanxdx=

A
π/4
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B
π/3
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C
0
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D
None of these
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Solution

The correct option is A π/4
I=π011+(tanx)1/4dx=π2011+(tanx)1/4dx
2I=π20(11+(tanx)1/4+(tanx)1/41+(tanx)1/4)dx=π20(1+(tanx)1/41+(tanx)1/4)dx
2I=π201dx2I=π2
I=π4

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