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Evaluate `int tan^(3)x dx` |
Answer» Correct Answer - `(1)/(2) tan^(2)x-log|sec x|+C` `I=int tan^(3)x dx` `=int tan^(2)x tanx dx` `=int (sec^(2)x-1)tanx dx` `=int tanx sec^(2) x dx - int tanx dx` `=I_(1)-log|secx|+C, " where " I_(1)=int tanx sec^(2)x dx` Putting `tan x =t " and " sec^(2) x dx=dt " in " I_(1),` we get `I_(1)= int t dt = (t^(2))/(2)=(1)/(2) tan^(2)x +C` Hence, `I=(1)/(2) tan^(2)x-log|sec x|+C`. |
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