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0π/2sin.xsinx+cosxdx=

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a
π4
b
π2
c
0
d
1

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detailed solution

Correct option is A

I=∫0π/2 sin⁡xsin⁡x+cos⁡xdx-----1I=∫0π/2 sin⁡π2−xsin⁡π2−x+cos⁡π2−xdx=∫0π/2 cos⁡xcos⁡x+sin⁡xdx-----2(1)+(2)⇒2I=∫0π/2 (1)dx=(x)0π/2=π2I=π4


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