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`int_(0)^(pi)(x sinx)/((1+sinx))dx=pi((pi)/(2)-1)` |
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Answer» `2I=pi*int_(0)^(pi)(sinx)/((1+cosx))dx=pi*int_(0)^(pi)20(1-(1)/(1+sinx))dx` `=pi^(2)-pi*int_(0)^(pi)((1-sinx)/(1-sin^(2)x))dx=pi^(2)-pi*int_(0)^(pi)((1-sinx))/(cos^(2)x)dx` `pi^(2)-pi*int_(0)^(pi)(sec^(2)x-secxtanx)dx`. |
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