How do you find the general antiderivative of #f(x)= sin2x+cos2x#?

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Answer 1 · 2015-02-25T19:42:42

The answer is: #-1/2cos2x+1/2sin2x+c#.

This is because:

#int(sin2x+cos2x)dx=1/2int(2sin2x+2cos2x)dx=#

#=-1/2cos2x+1/2sin2x+c#,

using these rules:

#intsinf(x)*f'(x)dx=-cosf(x)+c#

and

#intcosf(x)*f'(x)dx=sinf(x)+c#