Prove by induction or otherwise that $$\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2$$ for every integer $n\ge0$.
How to prove the above question?
Can it be proved without using induction?
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Well I imagine Weierstrass substitution should work, but its probably easier by induction and good practice. What have you tried? – Calvin Khor Apr 11 '14 at 08:34
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Maybe you could use expression of $\sin(2n+1)x/\sin x$ obtained from this question (or other questions linked there): http://math.stackexchange.com/questions/225941/proving-sum-limits-k-0n-coskx-frac12-frac-sin-frac2n12x – Martin Sleziak Apr 11 '14 at 12:43
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For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Apr 11 '14 at 12:50
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For $n=0$, the equality is trivial. Then note that: $$ \int_0^{\frac\pi2} \frac{\sin(2n+3)x}{\sin x}dx=\int_0^{\frac\pi2} \frac{\cos 2x\sin(2n+1)x}{\sin x}dx+\int_0^{\frac\pi2} \frac{\sin 2x\cos(2n+1)x}{\sin x}dx $$ And use $\cos 2x=1-2\sin^2x$ and $\sin2x=2\cos x\sin x$. You get: $$ \int_0^{\frac\pi2} \frac{\sin(2n+3)x}{\sin x}dx=\int_0^{\frac\pi2} \frac{\sin(2n+1)x}{\sin x}dx-2\int_0^{\frac\pi2} \cos(2n+2)x dx. $$ Now it is not difficult to see that $\int_0^{\frac\pi2} \cos(2n+2)x dx=0$. This will constitute the induction step.
Arash
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Let $$I_n = \int_0^{\pi/2} \dfrac{\sin(2n+1)x}{\sin(x)}dx$$ We then have $$I_{n+1} - I_n = \int_0^{\pi/2} \dfrac{2 \sin(x)\cos(2(n+1)x)}{\sin(x)} dx = 2\int_0^{\pi/2} \cos(2(n+1)x) dx = 0$$ for $n \neq -1$. Hence, $$I_n = I_0 = \dfrac{\pi}2$$ Essentially, we are saying that $\dfrac{dI_n}{dn} = 0$, where the derivative is to be interpreted in discrete sense and concluding that $I_n$ must be a constant.
user141421
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