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`int_(-(3pi)/2)^(-pi/2) {(pi+x)^3+cos^2(x+3pi)}dx` is equal to (A) `pi/4-1` (B) `pi^4/32` (C) `pi^4/32+pi/2` (D) `pi/2` |
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Answer» Correct Answer - `pi//2` Let `I=int_(-3pi//2)^(-pi//2)[(x+pi)^(3)+cos^(2)(x+3pi)]dx` Put `x=pi=t`, so that `dx=dt` when `x=-(3pi)/2, t=-(pi)/2` When `x=-(pi)/2,t=(pi)/2` `:.I=int_(-pi//2)^(pi//2) [t^(3)+cos^(2)(t+2pi)]dt` `=int_(-pi//2)^(pi//2) [t^(3)+cos^(2)t]dt` `=0+2int_(0)^(pi//2) cos^(2) t dt =int_(0)^(pi//2)(1+cos2t)dt` `=(pi)/2+0=(pi)/2` |
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