- #1
yungman
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I use part of HS-Scientist derivation in another post thanks to his detail derivation. I want to solving ## \int_0^{\pi} \sin(x sin(\theta)) d\theta##
[tex] \int_0^{\pi} \sin^m(\theta) d\theta=-\frac{1}{m}\left[sin^{m-1}(x)cos(\theta)\right]_0^{\pi}+\frac{m-1}{m} \int sin^{m-2}(\theta) d\theta= \frac{m-1}{m} \int sin^{m-2}(\theta) d\theta[/tex]
[tex]\hbox{As }\;\left[sin^{m-1}(x)cos(\theta)\right]_0^{\pi}=0[/tex]
Let ##m=2n+1##
[tex]\Rightarrow\; \int_0^{\pi} \sin^m(\theta) d\theta=\int_0^{\pi} \sin^{2n+1}(\theta) d\theta=\frac{2n}{2n+1}\int_0^{\pi} \sin^{2n-1}\theta d\theta=\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\int_0^{\pi}\sin\theta d\theta[/tex]
[tex]\Rightarrow\; \int_0^{\pi} \sin^{2n+1}(\theta) d\theta=\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\int_0^{\pi}\sin\theta d\theta =2\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}[/tex]
[tex]\sin x =\sum_0^{\infty}\frac {(-1)^n x^{2n+1}}{(2n+1)!}[/tex]
[tex]\Rightarrow\;\int_0^\pi \sin(x sin(\theta)) d\theta=\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!} \int_0^{\pi}\sin^{2n+1}\theta d\theta\;=\;\left[\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!}\right]\left[2\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\right][/tex]
[tex]\Rightarrow\;\int_0^\pi \sin(x sin(\theta)) d\theta \;=\;\sum_0^{\infty} \frac {(-1)^n 2x^{2n+1}}{(2n+1)!} \frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))} \;=\;\;\sum_0^{\infty} \frac {(-1)^n 2x^{2n+1}}{3^2 \cdot 5^2\cdot 7^2\cdot \cdot \cdot \cdot (2n+1)^2}[/tex]
I want to verify I am correct in my derivation.
Thanks
[tex] \int_0^{\pi} \sin^m(\theta) d\theta=-\frac{1}{m}\left[sin^{m-1}(x)cos(\theta)\right]_0^{\pi}+\frac{m-1}{m} \int sin^{m-2}(\theta) d\theta= \frac{m-1}{m} \int sin^{m-2}(\theta) d\theta[/tex]
[tex]\hbox{As }\;\left[sin^{m-1}(x)cos(\theta)\right]_0^{\pi}=0[/tex]
Let ##m=2n+1##
[tex]\Rightarrow\; \int_0^{\pi} \sin^m(\theta) d\theta=\int_0^{\pi} \sin^{2n+1}(\theta) d\theta=\frac{2n}{2n+1}\int_0^{\pi} \sin^{2n-1}\theta d\theta=\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\int_0^{\pi}\sin\theta d\theta[/tex]
[tex]\Rightarrow\; \int_0^{\pi} \sin^{2n+1}(\theta) d\theta=\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\int_0^{\pi}\sin\theta d\theta =2\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}[/tex]
[tex]\sin x =\sum_0^{\infty}\frac {(-1)^n x^{2n+1}}{(2n+1)!}[/tex]
[tex]\Rightarrow\;\int_0^\pi \sin(x sin(\theta)) d\theta=\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!} \int_0^{\pi}\sin^{2n+1}\theta d\theta\;=\;\left[\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!}\right]\left[2\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\right][/tex]
[tex]\Rightarrow\;\int_0^\pi \sin(x sin(\theta)) d\theta \;=\;\sum_0^{\infty} \frac {(-1)^n 2x^{2n+1}}{(2n+1)!} \frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))} \;=\;\;\sum_0^{\infty} \frac {(-1)^n 2x^{2n+1}}{3^2 \cdot 5^2\cdot 7^2\cdot \cdot \cdot \cdot (2n+1)^2}[/tex]
I want to verify I am correct in my derivation.
Thanks
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