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The value of `int_-pi^pi cos^2x/[1+a^x].dx`,a>0 isA. `pi`B. `api`C. `(pi)/(2)`D. `2pi` |
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Answer» Correct Answer - C Let `I=int_(-pi)^(pi)(cos^(2)x)/(1+a^(x))dx` . . . (i) `=int_(pi)^(-pi)(cos^(2)(-x))/(1+a^(-x))d(-x)` `rArrI=int_(-pi)^(pi)a^(x)(cos^(2)x)/(1+a)dx` On adding Eqs . (i) and (ii) , we get `2I=int_(-pi)^(pi)((1+a^(x))/(1+a^(x)))cos^(2)xdx` `=int_(-pi)^(pi)cos^(2)xdx=2int_(0)^(pi)(1+cos2x)/(2)dx` `=int_(0)^(pi)(1+cos2x)dx` `=int_(0)^(pi)1dx+int_(0)^(pi)cos2xdx` `=[x]_(0)^(pi)+2int_(0)^(pi//2)cos2xdx=pi+0` `rArr2I=pirArrI=pi//2` |
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