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Q.

∫−ππcos2x1+axdx  where a>0  is

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a

π2

b

c

d

πa

answer is A.

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Detailed Solution

∫−bbf(x)dx=∫0b[f(x)+f(−x)]dx→ ∫−ππcos2x1+axdx=∫0π(cos2x1+ax+cos2x1+a−x)dx                          =∫0πcos2xdx                        =2∫0π2sin2xdx=2⋅12⋅π2=π2
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