What is the antiderivative of $- sin x d x$ $-sinx dx$ ?

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What is the antiderivative of $- sin x d x$ $-sinx dx$ ?

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Answer 1 · 2015-02-26T04:34:31

Let I = $\int - sin x . d x$ $int -sinx.dx$
I = $- \int sin x . d x$ $-int sinx.dx$
I = $- (- cos x)$ $- (-cos x)$

(Since $\frac{d}{d x} (cos x) = - sin x$ $d/dx (cosx)= -sinx$ ,
$- \frac{d}{d x} (cos x) = sin x$ $-d/dx (cosx) = sinx$ )

Thus, I = cos x

Note that if you remember $\frac{d}{d x} (cos x) = - sin x$ $d/dx (cosx)= -sinx$ , you can directly get the answer since the antiderivative (integral) of a derivative is nothing but the function which was differentiated.

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What is the antiderivative of $- sin x d x$ $-sinx dx$ ?

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