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Evaluate: `int_(-pi//2)^(pi//2)(cosx)/(1+e^x)dx`A. `(pi^(2))/4-2`B. `(pi^(2))/4+2`C. `pi^(2)-e^((pi)/2)`D. `pi^(2)+e^(pi)/2` |
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Answer» Correct Answer - A `I=int_(-(pi)/2)^((pi)/2)(x^(2)cosx)/(1+e^(x))dx`………………1 `=int_(-(pi)/2)^((pi)/2)(x^(2)cosx)/(1+1/(e^(x))`(Replacing x by `-x`) `:. I=int_(-(pi)/2)^((pi)/2)(x^(2)cosx.e^(x))/(1+e^(x))dx`……………..2 Adding 1 and 2 we get `2I=int_(-(pi)/2)^((pi)/2)x^(2)cosx dx` `:. I=int_(0)^((pi)/2)x^(2)cosxdx` `=x^(2). sin x|_(0)^(pi//2)-int_(0)^((pi)/2)2x sin xdx` `=(pi^(2))/4-2[(-xcosx)_(0)^(pi//2)-int_(0)^(pi//2)( -cosx)dx]` `=(pi^(2))/4-2[1]=(pi^(2))/4-2` |
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