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Question

π23π2[(x+π)3+cos2(x+π)]dx is equal to

A
(π432)+(π2)
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B
π2
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C
1
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D
π432
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Solution

The correct option is B π2
Let l=π23π2[(x+π)3+cos2(x+π)]dx
Put x+π=t
l=π2π2[t3+cos2t]dt
l=π2π2cos2t dt=π2

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