alleryFind $\displaystyle\int \sin x\,dx$.
A$\cos x + C$
B$-\cos x + C$
C$-\sin x + C$
D$\sin x + C$
Answer & Solution
Correct answer: B. $-\cos x + C$
$\dfrac{d}{dx}(-\cos x) = \sin x$, so $\displaystyle\int \sin x\,dx = -\cos x + C$.
Mnemonic: differentiating gives a negative on the sine term ($\dfrac{d}{dx} \cos x = -\sin x$). To reverse, you keep the sign that makes the derivative *positive* $\sin x$, which is the $-\cos x$ choice.
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