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Evaluate:`int_(-pi/2)^(pi/2)log{(a x^2+b x+c)/(a x^2-b x+c)(a+b)|sinx|}dx` |
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Answer» `I=int_(-pi//2)^(pi//2) log {(ax^(2)+bx+c)/(ax^(2)-bx+c)(a+b)|sinx|}dx` `=int_(-pi//2)^(pi//2) log((ax^(2)+bx+c)/(ax^(2)-bx+c))dx+int_(-pi//2)^(pi//2) log (a+b)dx` `+int_(-pi//2)^(pi//2) log|sinx|dx`……………1 `=I_(1)+I_(2)+I_(3)` Now let `f(x)=log((ax^(2)+bx+c)/(ax^(2)-bx+c))` or `f(-x)=log((ax^(2)-bx+c)/(ax^(2)+bx+c))` `=-f(x)` `:.I_(1)=int_(-pi//2)^(pi//2) f(x)dx=0` `I_(2)=log(a+b)[x]_(-pi//2)^(pi//2)` `=pi log(a+b)` `I_(3)=int_(-pi//2)^(pi//2) log|sinx|dx` `=2int_(0)^(pi//2)log|sinx|dx` `=2int_(0)^(pi//2)logsinx dx` `=2(-1/2pi log 2)` Hence, from 1 we have `I=0+pi log(a+b)-pi log 2` `=pi log{(a+b)//2}` |
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