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Question

The value of π402tan3x dx is


A
ln2
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B
ln21
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C
π4
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D
π2
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Solution

The correct option is C ln21
I=π402(1sec2x)tanx dx
I=π402tanx dxπ402tanxsec2x dx
I=2ln|secx|π40tan2xπ40
I=2ln|2|2ln|1|(10)
I=ln21

Mathematics

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