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`int ((1+tanx)/(1-tanx))dx` |
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Answer» Correct Answer - `-log |cosx 0 sinx| + C` `I = int((1+(sinx)/(cosx))/(1-(sinx)/(cosx)))dx = int((cosx+sinx))/((cosx-sinx)) dx` `= -int(dt)/(t)`, where `(cosx -sinx) =t` `= - log|t| +C =-log|cosx-sinx|+C`. |
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