Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

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 $$\left[\dfrac {x}{2}-\dfrac{\sin 2Cx}{4C}\right]_0^d$$ $$\dfrac {d}{2}-\dfrac{\sin 2Cd}{4C}$$

share|improve this answer
1  
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

Hints:

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

$$\int\sin^2xdx=\frac12\int(1-\cos2x)dx=\frac12\left(x-\frac12\sin2x\right)$$

Finally, do a simple substitution

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.