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0π/2sinxsinx+cosxdx is equal to

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a
0
b
π4
c
π2
d
None of these

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detailed solution

Correct option is B

Firstly, reduce the integrand into simplest form by usingthe property ∫0a f(x)dx=∫0a f(a−x)dx add them and integrate.LetI=∫0π/2 sin⁡xsin⁡x+cos⁡xdx---ithen    I=∫0π/2 sin⁡π2−xsin⁡π2−x+cos⁡π2−xdx      ∵∫0a f(x)dx=∫0a f(a−x)dx⇒I=∫0π/2 cos⁡xcos⁡x+sin⁡xdx-----ii∵sin⁡π2−x=cos⁡x and cos⁡π2−x=sin⁡xOn adding Eqs. (i) and (ii), we get2l=∫0π/2 sin⁡x+cos⁡xsin⁡x+cos⁡xdx=∫0π/2 1dx=[x]0π/2=π2−0⇒ I=π4


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