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Q.

∫e−1tanxtdt1+t2+∫e−1cotxdtt(1+t2)=

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a

1

b

e

c

e−1

d

π4

answer is A.

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Detailed Solution

Setting t=1u  in the second integral, it becomes ∫tanxeudu1+u2 ∴I=∫e−1tanxtdt1+t2+∫tanxetdt1+t2=∫e−1etdt1+t2=12ln(1+t2)|1ee                                                 =12ln(1+e21+e−2)=12lne2=1
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