Question Number 29506 by abdo imad last updated on 09/Feb/18
$${le}\:{give}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sin}\left(\left(\mathrm{2}{n}−\mathrm{1}\right){x}\right)}{{sinx}}{dx}\:{and}\:{B}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sin}^{\mathrm{2}} \left({nx}\right)}{{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\left.\mathrm{1}\right){calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:{B}_{{n}+\mathrm{1}} −{B}_{{n}} =\:{A}_{{n}+\mathrm{1}} .{then}\:{find}\:{B}_{{n}} . \\ $$
Commented by abdo imad last updated on 11/Feb/18
![A_(n+1) −A_n = ∫_0 ^(π/2) ((sin(2n+1)x −sin(2n−1)x)/(sinx))dx but sin(2n+1)x= sin(2nx)cosx +cos(2nx)sinx sin(2n−1)x=sin(2nx)cosx −cos(2nx)sinx ⇒ sin(2n+1)x −sin(2n−1)x=2cos(2nx)sinx⇒ A_(n+1) −A_(n ) = ∫_0 ^(π/2) 2cos(2nx)dx=[(1/n)sin(2nx)]_0 ^(π/2) =0 ∀n A_n =A_1 = (π/2) .](Q29730.png)
$${A}_{{n}+\mathrm{1}} −{A}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sin}\left(\mathrm{2}{n}+\mathrm{1}\right){x}\:−{sin}\left(\mathrm{2}{n}−\mathrm{1}\right){x}}{{sinx}}{dx}\:{but} \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{1}\right){x}=\:{sin}\left(\mathrm{2}{nx}\right){cosx}\:+{cos}\left(\mathrm{2}{nx}\right){sinx} \\ $$$${sin}\left(\mathrm{2}{n}−\mathrm{1}\right){x}={sin}\left(\mathrm{2}{nx}\right){cosx}\:−{cos}\left(\mathrm{2}{nx}\right){sinx}\:\Rightarrow \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{1}\right){x}\:−{sin}\left(\mathrm{2}{n}−\mathrm{1}\right){x}=\mathrm{2}{cos}\left(\mathrm{2}{nx}\right){sinx}\Rightarrow \\ $$$${A}_{{n}+\mathrm{1}} −{A}_{{n}\:} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{2}{cos}\left(\mathrm{2}{nx}\right){dx}=\left[\frac{\mathrm{1}}{{n}}{sin}\left(\mathrm{2}{nx}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{0}\:\forall{n} \\ $$$${A}_{{n}} ={A}_{\mathrm{1}} =\:\frac{\pi}{\mathrm{2}}\:. \\ $$