Simple u-subsitution - Paradoxical Result

Question

If I were to try and take $$\int{\mathrm{sin}(t)\mathrm{cos}(t)dt} $$ I would either take $u=\mathrm{sin}(t) $, yeilding a result of $\frac{1}{2} \mathrm{sin}^2(t) + C$, or I would take $u=\mathrm{cos}(t) $, yeilding a result of $-\frac{1}{2} \mathrm{cos}^2(t)+ C$.

These two results are not equivalent. What just happened?

$1/2 \sin^2 t=1/2-1/2\cos^2 t$. And don't forget that we need always to add a constant. — Artem, May 03 '13 at 00:21
Does this answer your question? A contradictory integral: $\int \sin x \cos x , \mathrm dx$ — Hans Lundmark, Apr 01 '21 at 14:23

score 4 · Answer 1 · edited May 03 '13 at 00:27

4

The answers are not equal, but they are equivalent. Remember, as Artem said, that the two answers differ by a constant. And antiderivatives are always defined up to a constant.

edited May 03 '13 at 00:27

DonAntonio

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answered May 03 '13 at 00:24

Ted Shifrin

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Joseph G. · Accepted Answer · 2013-05-03T00:32:20.130

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If you evaluate the definite integral of $$ \int_{a}^{b} \sin(t)\cos(t) dt$$, then you get:$$\frac{1}{2}\left(\sin^2(b)-\sin^2(a)\right)=\frac{1}{2}\left(1-\cos^2(b)-1+\cos^2(a)\right)$$ $$=-\frac{1}{2}\left(\cos^2(b)-\cos^2(a)\right).$$ So while the indefinite integral might seem off, the definite integral has the same values.

edited May 03 '13 at 00:32

answered May 03 '13 at 00:22

Joseph G.

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Every time, despite the fact that their graphs are not equivalent? Hot dog... – user1833028 May 03 '13 at 00:23
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The fact that their graphs are only off by a constant implies that their derivatives should be equal. – Joseph G. May 03 '13 at 00:26
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Wait I get it lol. – user1833028 May 03 '13 at 00:27
Im just showing that regardless of which antiderivative you use, either the $\frac{1}{2}\sin^2(t)$ or the $-\frac{1}{2}\cos^2(t)$, you will still get the same answer. – Joseph G. May 03 '13 at 00:30

Simple u-subsitution - Paradoxical Result

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