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I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different?

$ \int \sin x \cos x dx $

1) via subsitution $ u = \sin x $ $ u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Rightarrow \int \sin x \cos x dx = \frac12 \sin^2 x $

2) via subsitution $ u = \cos x $ $ u = \cos x; du = -\sin x dx \Rightarrow -\int udu = -\frac12 u^2 \Rightarrow \int \sin x \cos x dx = -\frac12 \cos^2 x = -\frac12 (1 - \sin^2 x) = -\frac12 + \frac12 \sin^2 x $

3) using $ \sin 2x = 2 \sin x \cos x $

$ \int \sin x \cos x dx = \frac12 \int \sin 2x = \frac12 (- \frac12 \cos 2x) = - \frac14 \cos 2x = - \frac14 (1 - 2 \sin^2 x) = - \frac14 + \frac12 \sin^2 x $

So, we have: $$ \frac12 \sin^2 x \neq -\frac12 + \frac12 \sin^2 x \neq - \frac14 + \frac12 \sin^2 x $$

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    $\begingroup$ Ahh! I always give this problem in my first year calculus course. $\endgroup$ – Jyrki Lahtonen May 4 '13 at 14:40
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    $\begingroup$ $+C$... ${}{}{}$ $\endgroup$ – David Mitra May 4 '13 at 14:40
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    $\begingroup$ $*$ is usually used for convolution in this context. I removed it. $\endgroup$ – Ayman Hourieh May 4 '13 at 14:42
  • $\begingroup$ @AymanHourieh: Didn't I? $\endgroup$ – Inceptio May 4 '13 at 14:44
  • $\begingroup$ @Inceptio Check the edit history. I edited the body; you edited the title. $\endgroup$ – Ayman Hourieh May 4 '13 at 14:49
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Antiderivatives are only unique up to adding a constant ('of integration'). If you were to stick limits in your integrals then you'd always get the same number.

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  • $\begingroup$ Oh, yes. I checked it with 0 and \pi/2 and those 'strange' fractions deducted each other. Thank you $\endgroup$ – Jimch May 4 '13 at 14:46
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Note: You are calculating indefinite integral and constants can be anything(they may differ). In fact the general solution to that would be just $C+\dfrac{\sin^2 x}{2}$

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$$\frac{d\{f(x)+c\}}{dx}=f'(x)$$ for any arbitrary constant $c$

$$\implies \int f'(x)dx=f(x)+d $$ for any arbitrary constant $d$

So, in indefinite integral we can get answers which differ by some constant

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A primitive is unique up to a constant

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