if you are integrating with respect to x, the indefinite integral of 1 is just x
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By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
-1/15sin5ycos3y
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Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C