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I can get the solution but somewhere I am going wrong. Can't figure out where. We have to solve: $$ \int -\sin(-x+2) \, dx$$

Here is my step by step solution using $u$-substitution:

$$ u = -x + 2 \iff du = -dx $$

$$ \int -\sin(-x+2) \, dx \iff \int \sin(-x+2) \, (-dx ) \iff \int \sin(u) \, du$$

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

$$ \int \sin(u) \, du \iff -\cos(u) + C \iff -\cos(-x+2) + C $$

Which seems to be the solution. Now if I differentiate it then I should get back the integral but I don't.

$$ \frac{d}{dx} (-\cos(-x+2) + C) \iff -\sin(-x+2) \, (-1) $$

$ \sin(-x+2) $ which is not equal to where we started from $-\sin(-x+2)$. Am I missing something ?

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  • $\begingroup$ when you are taking derivative in the last step, there will be one more $-$ sign. Derivative of cosine is $-\sin$. $\endgroup$ – Anurag A Sep 14 '18 at 5:12
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    $\begingroup$ $\frac{d}{dx}\left(-\cos(-x + 2)\right) = -\frac{d}{dx}\left(\cos(-x + 2)\right) = -(-\sin(-x+2)(-1)) = -\sin(-x+2)$ $\endgroup$ – Lucas Corrêa Sep 14 '18 at 5:13
  • $\begingroup$ may be I am working too much and need a break. Just missed a simple thing :( $\endgroup$ – Arnuld Sep 14 '18 at 5:18
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Your calculation of the indefinite integral is actually correct, and you just dropped a negative sign when calculating the derivative. As you said,

$$\frac{d}{dx}\cos(x) = -\sin(x),$$

so

\begin{align} & \frac{d}{dx}[-(\cos(-x-2) +C] = -(-\sin(-x+2)(-1)) \\[10pt] = {} & -(\sin(-x+2)) = -\sin(-x+2) \end{align}

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  • $\begingroup$ Proper notation is $\sin(x)$ or $\sin x,$ not $sin(x).$ I edited accordingly. $\endgroup$ – Michael Hardy Sep 14 '18 at 16:27
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$$I=\int\sin(-x+2)dx$$ $u=-x+2$ so $dx=-du$ $$I=-\int\sin(u)du=\cos(u)+C=cos(-x+2)+C$$ now $$\frac{d}{dx}\left[\cos(-x+2)\right]=-\sin(-x+2)*\frac{d}{dx}\left[-x+2\right]=\sin(-x+2)$$

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