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∫0π2/4cosx dx

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a

π-2

b

π+2

c

π2−1

d

2

answer is A.

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Detailed Solution

Put x=t⇒x=t2⇒dx=2tdt∫0π2/4 cos⁡xdx=∫0π/2 (cos⁡t)⋅2tdt LL⇒ t=0,U.Lt=π24⇒t=π2 = 2t sint π20 -∫0π22sint dt =π-2

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