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`int_(0)^(pi//2)log(tanx+cotx)dx=pi(log2)` |
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Answer» `I=int_(0)^(pi//2)log((sinx)/(cosx)+(cosx)/(sinx))dx=int_(0)^(pi//2)log((1)/(sinxcosx))dx` `=-[int_(0)^(pi//2)log(sinx)dx+int_(0)^(pi//2)log(cosx)dx]`. |
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