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## Homework Statement

The actual question is asking for the normalization constant for the wavefunction

[Tex] \psi\left(x\right)=A\sin^{5}\left(\dfrac{\pi x}{a}\right) [/Tex]

**without carrying out integration**

In short they want me to find the value of A such that

[Tex] A^{2}\int_{0}^{a}\sin^{10}\left(\dfrac{\pi x}{a}\right)=1 [/Tex]

## Homework Equations

## The Attempt at a Solution

I know the integral comes out nicely to [TeX] 63a/256 [/TeX] so [TeX] \sqrt{\dfrac{256}{63a}} [/TeX]

I figure that there must be some linear map from the width of the 'half cycle' to the area under the graph. So we are taking [TeX] \dfrac{a}{ \pi } [/TeX] of the area

[Tex] \int_{0}^{pi}\sin^{10}\left(x\right) [/Tex]

What remains is for me to actually show that

[Tex] \int_{0}^{pi}\sin^{10}\left(x\right)=\dfrac{63 \pi}{256} [/Tex]

Is there some method of series expansion here I can use to show this? I think something's going to happen to take out all 2nd, 3rd, 4th... terms of the series expansion because of the [Tex]2n \pi [/Tex] cycles or something along those lines.

Or is there an entirely different way to find the area under the curve without integrating that I missed out? Thanks for your help!