integrate sin(x)cos(x) using trig identity.

Question

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 $\frac{1}{2}\sin^2(x)$?

Answer 1

Because $$\frac{1}{2} \sin^2(x) = \frac{1-\cos(2x)}{4}$$

Answer 2

HINT

The answers are equivalent. Recall that $$ \cos (2x) = \cos^2 x - \sin^2 x = \left(1-\sin^2 x\right) - \sin^2 x = 1 - 2 \sin^2x $$

Answer 3

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.

Answer 4

Observe that $\int \sin 2x\ \text {dx} = -\frac {\cos 2x} {2} + C$ and use the fact that $\cos 2x =1-2{\sin}^2 x.$

Answer 5

To answer the actual question (“How did the book”, etc.): probably just by inspection (chain rule backwards), or else via the substitution $u=\sin x$, $du = \cos x \, dx$.