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Math Help - integration question no. 7

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    integration question no. 7

    integration question no. 7
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    \displaystyle I_n=\int_0^{\pi}\frac{\sin (2n-1)x}{\sin x}dx=\int_0^{\pi}\frac{\sin 2nx\cos x-\sin x\cos 2nx}{\sin x}dx=
    \displaystyle=\int_0^{\pi}\frac{\sin 2nx\cos x}{\sin x}dx-\int_0^{\pi}\cos 2nxdx=
    \displaystyle=\frac{1}{2}\int_0^{\pi}\frac{\sin(2n  +1)x+\sin(2n-1)x}{\sin x}dx-\left.\frac{1}{2n}\sin 2nx\right|_0^{\pi}=
    \displaystyle=\frac{1}{2}I_{n+1}+\frac{1}{2}I_n\Ri  ghtarrow\frac{1}{2}I_n=\frac{1}{2}I_{n+1}\Rightarr  ow I_n=I_{n+1}
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  3. #3
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    Hello, afeasfaerw23231233!

    Let: . I_n \;=\;\int^{\pi}_0\frac{\sin(2n+1)x}{\sin x}\,dx

    Bu using the identity: . \sin(A - B) \:=\:\sin A\cos B - \sin B\cos A

    . . show that: . I_{n+1} \:=\:I_n
    I will drop the limits (for now) . . .


    We have: . I_n \;=\;\int \frac{\sin(2nx - x)}{\sin x}\,dx \;=\;\int\frac{\sin(2nx)\cos(x) - \sin(x)\cos(2nx)}{\sin x}\,dx

    . . I_n \;=\;\int\left[\frac{\sin(2nx)\cos(x)}{\sin(x)} - \cos(2nx)\right]\,dx \;=\;\int\frac{\sin(2nx)\cos(x)}{\sin(x)}\,dx \,- \int \cos(2nx)\,dx


    We have: . I_{n+1} \;=\;\int\frac{\sin(2nx + x)}{\sin x}\,dx \;=\;\int\frac{\sin(2nx)\cos(x) + \sin(x)\cos(2nx)}{\sin x}\,dx

    . . I_{n+1} \;=\;\int\left[\frac{\sin(2nx)\cos(x)}{\sin x} + \cos(2nx)\right]\,dx \;= . \int\frac{\sin(2nx)\cos(x)}{\sin x}\,dx + \int \cos(2nx)\,dx



    We see that I_n and I_{n+1} are identical except for their second integrals.

    . . But: . \int^{\pi}_0\cos(2nx)\,dx \;=\;\frac{1}{2n}\sin(2nx)\,\bigg|^{\pi}_0\;=\;\fr  ac{1}{2n}\left[\sin(2n\pi) - \sin(0)\right] \;=\;0


    Therefore: . I_{n+1} \;=\;\int^{\pi}_0\frac{\sin(2xn)\cos(x)}{\sin x}\,dx \;=\;I_n

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