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I feel like there is something I am missing here. When integrating both sides of the trigonometric identity $\sin{2x}=2\cos x\sin x$ I get different results.

The left side of course results in $-\frac{1}{2}\cos{2x}+C$.

The right side I solve with u-substitution:

$u=\cos x$

$du=-\sin x dx$

$-2\int udu=-u^2+C=-\cos^2 x+C$

While writing this question I noticed another identity $\cos^2 x=\frac{1}{2}+\frac{1}{2}\cos 2x$. So apparently the $\frac{1}{2}$ falls out because of the $+C$ resulting from indefinite integration? This is still a little confusing to me.

Rubenknex
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2 Answers2

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You are correct to recall that $\cos^2x=\frac{1+\cos 2x }{2}$.

This is an indefinite integral. So, the constant term $\frac 12$ is not relevant. That is

$$\int \sin 2x \,dx=-\frac12\cos (2x)+C_1 \tag 1$$

and

$$\begin{align} \int \sin 2x \,dx&=-\cos^2 x+C_2\\\\ &=-\frac12\cos 2x+(-\frac12 +C_2)\\\\ &=-\frac12\cos 2x+C_3\tag 2 \end{align}$$

where we absorbed the constants $-\frac12+C_2$ into a new constant and called that new constant $C_3$. Inasmuch as the integration constant is arbitrary, $(1)$ and $(2)$ are equivalent statements.

Mark Viola
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Integration process gives you results correct up to an arbitrary constant.

Both results are essentially the same.

$$ - \frac{\cos 2 x}{2} + C_1 = - \frac { 2 \cos ^2 x -1}{2} + C_1 = - \cos ^2 x + C_2 $$

That is why it is better to write $ C_1, C_2 $ for the integration constants.

Narasimham
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