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I've been thinking about integration lately, and I've come up with a question that I'm not sure how to address. Consider $$ \int \sin x\cos x \, \mathrm dx = - \int -\sin x \cos x \, \mathrm dx $$ I started with the integral on the left hand side, which suggests a typical $u$-substitution. Let $u=\sin x$ then $\, \mathrm du=\cos x \, \mathrm dx$. So the integral evaluates to $$ \int \sin x\cos x \, \mathrm dx = \frac{\sin^2(x)}{2} $$ But the original integral also suggests an alternate substitution. Let $u=\cos x$ and then $du=-\sin x \, dx$. So now $$ \int \sin x\cos x \, \mathrm dx =- \int -\sin x \cos x \, \mathrm dx= -\frac{\cos^2(x)}{2} $$ So now I have that the integral evaluates to two different functions. I've tried playing with some different trigonometric identities, but I haven't been able to show that this is true and I'm fairly certain I haven't had any success because the statement itself isn't true. What am I doing wrong? How do you evaluate $\int \sin x\cos x \, \mathrm dx$?

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The only thing that you’re doing wrong is forgetting the constant of integration: your two antiderivatives differ by a constant, so of course they are antiderivatives of the same function.

$$\frac12\sin^2x=-\frac12\cos^2x+\frac12$$

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  • $\begingroup$ OH. Obviously. Thank you for your quick response. $\endgroup$ – chris Nov 10 '12 at 3:58
  • $\begingroup$ @chris: You’re welcome. $\endgroup$ – Brian M. Scott Nov 10 '12 at 3:59

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