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0π/2dx2cosx+3 is equal to

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a
25tan−1⁡15
b
15tan−1⁡15
c
1
d
23tan−1⁡15

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detailed solution

Correct option is A

Put tan⁡x/2=t The given integral reduces to ∫01 2dt1+t221−t21+t2+3=2∫01 dt2−2t2+3+3t2=2∫01 dtt2+5                           =25tan−1⁡t501=25tan−1⁡15


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