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Evaluate `int_(0)^(2pi) (dx)/(1+3cos^(2)x)` |
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Answer» `I=int_(0)^(2pi)(dx)/(1+3cos^(2)x)` `=int_(0)^(2pi)(sec^(2) xdx)/(sec^(2)x+3)` `=int_(0)^(2pi)(sec^(2) xdx)/(4+tan^(2)x)` Here we cannot substitute `tan x=t` as `tanx` is discontinuous in `(0,2pi)` Let `f(x)=(sec^(2)x)/(4+tan^(2)x)` `:f(2pix)=f(x)` `:.I=2 int_(0)^(pi)(sec^(2) xdx)/(4+tan^(2)x)` Also `f(pi-x)=f(x)` `:. I=4 int_(0)^(pi//2)(sec^(2) xdx)/(4+tan^(2)x` `=4 int_(0)^(oo) (dt)/(4+t^(2))` (Putting `tanx=t` as `tanx` is continuous in `(0,pi//2)` `=4/2[tan^(-1)t/2]_(0)^(oo)` `=2[ta ^(-1)oo-tan^(-1)0]` `=pi` |
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