How do I find the general antiderivative of $f (x) = {sin}^{2} x + {cos}^{2} x$ $f(x)= sin^2x + cos^2x$ ?

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How do I find the general antiderivative of $f (x) = {sin}^{2} x + {cos}^{2} x$ $f(x)= sin^2x + cos^2x$ ?

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Answer 1 · 2015-03-06T16:09:16

Actually ${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x)=1$
So your antiderivative is:
$\int [{sin}^{2} (x) + {cos}^{2} (x)] d x = \int 1 d x = x + c$ $int[sin^2(x)+cos^2(x)]dx=int1dx=x+c$

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How do I find the general antiderivative of $f (x) = {sin}^{2} x + {cos}^{2} x$ $f(x)= sin^2x + cos^2x$ ?

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How do I find the general antiderivative of f(x)=sin2x+cos2xf(x)= sin^2x + cos^2x?

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How do I find the general antiderivative of $f (x) = {sin}^{2} x + {cos}^{2} x$ $f(x)= sin^2x + cos^2x$ ?