Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$ \sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B) $$
$$ \begin{split} \int \sin(x)\cos(x) dx &= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin(x+x)\right) dx \\ &= \int \left(\frac{1}{2}\sin(0) + \frac{1}{2}\sin(2x)\right) dx \\ &= \int \left(\frac{1}{2}\sin(2x)\right) dx \\ &= -\frac{1}{2} \frac{1}{2}\cos(2x) +C\\ &= -\frac{1}{4} \cos(2x) +C \end{split} $$
How did the book arrive at the answer $\frac{1}{2}\sin^2(x)$?
$\endgroup$ 25 Answers
$\begingroup$Note that$$\cos(2x)=1-2\sin^2x$$From here you get$$-\frac{1}{4}\cos(2x) + C = -\frac{1}{4} + \frac{1}{2}\sin^2 x+C = \frac{1}{2}\sin^2x + C_1$$where $C_1 = -\frac{1}{4}+C$ is a new constant.
$\endgroup$ $\begingroup$HINT
The answers are equivalent. Recall that$$ \cos (2x) = \cos^2 x - \sin^2 x = \left(1-\sin^2 x\right) - \sin^2 x = 1 - 2 \sin^2x $$
$\endgroup$ $\begingroup$Because$$\frac{1}{2} \sin^2(x) = \frac{1-\cos(2x)}{4}$$
$\endgroup$ $\begingroup$Observe that $\int \sin 2x\ \text {dx} = -\frac {\cos 2x} {2} + C$ and use the fact that $\cos 2x =1-2{\sin}^2 x.$
$\endgroup$ $\begingroup$To answer the actual question (“How did the book”, etc.): probably just by inspection (chain rule backwards), or else via the substitution $u=\sin x$, $du = \cos x \, dx$.
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