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In the book of quantum mechanics I came across an integral which was supposed to be from a manual ($C$ is a constant):

\begin{align} \int\limits_{0}^d \sin^2\left( C x \right)\, d x = \left.\left(\frac{x}{2}- \frac{\sin(2Cx)}{4C}\right)\right|_0^d \end{align}

Where can I read more about this? I would be glad if anyone could provide me a proof.

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If you are looking for a quick and dirty proof, you can use the Fundamental Theorem of Integral Calculus and differentiate both sides with respect to $d$. Then, only a trigonometric identity is needed to conclude. – Lord_Farin Jun 19 '13 at 13:14
Good point. Thank you! – 71GA Jun 19 '13 at 13:20
up vote 4 down vote accepted

$$\sin ^2 Cx=\dfrac{1-\cos 2Cx}{2}$$ $$\int \sin ^2 Cx\;dx=\int\dfrac{1-\cos 2Cx}{2}\;dx$$ $$\int\dfrac 12-\dfrac{\cos 2Cx}{2}dx$$ $$\dfrac {x}{2}-\dfrac{\sin 2Cx}{2\cdot 2C}$$ $$\dfrac {x}{2}-\dfrac{\sin 2Cx}{4C}$$ Now just put given limits $$\left[\dfrac {x}{2}-\dfrac{\sin 2Cx}{4C}\right]_0^d$$ $$\dfrac {d}{2}-\dfrac{\sin 2Cd}{4C}$$

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Why did you switch $C$ in the question to $n$ here? – Thomas Andrews Jun 19 '13 at 13:20
@ThomasAndrews I forgot. – iostream007 Jun 19 '13 at 13:49


$$\sin^2x=\frac{1-\cos 2x}2$$


Finally, do a simple substitution

$$u=Cx\implies\;dx=\frac1Cdu\;,\;\;x=0\implies u=0\;,\;\;x=d\implies u=Cd\ldots\ldots$$

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