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Q.

∫π23π2[2sinx]dx=

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a

−π

b

0

c

−π2

d

π2

answer is C.

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Detailed Solution

x=π2+t→I=∫2π[2cost]dt                            =∫0π31dt+∫π3π20⋅dt+∫π22π3(−1)dt+∫2π3π(−2)dt                           =π3−π6−2π3=−π2
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