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Question Number 48494 by maxmathsup by imad last updated on 24/Nov/18
find A_n =∫_0 ^(π/2) ((1−cos(n+1)x)/(2sin((x/2))))dx .
$${find}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}−{cos}\left({n}+\mathrm{1}\right){x}}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18
A_n −A_(n−1) =∫_0 ^(π/2) ((cos(n)x−cos(n+1)x)/(2sin((x/2))))dx =∫_0 ^(π/2) ((2sin(2n+1)(x/2).sin((x/2)))/(2sin((x/2))))dx =∫_0 ^(π/2) sin(2n+1)(x/2)dx =(((−1)/(2n+1)))∣cos(2n+1)(x/2)∣_0 ^(π/2) =(((−1)/(2n+1))){cos(2n+1)(π/4)−1} =(1/(2n+1)){1−cos(n(π/2)+(π/4))} if n=even =(1/(2n+1))(1+(1/( (√2)))) or (1/(2n+1))(1−(1/( (√2)))) if n odd =(1/(2n+1))(1+(1/( (√2)))) or (1/(2n+1))(1−(1/( (√2)))) now considering A_n −A_(n−1) =(1/(2n+1))(1+(1/( (√2)))) pls wait....
$${A}_{{n}} −{A}_{{n}−\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({n}\right){x}−{cos}\left({n}+\mathrm{1}\right){x}}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{{x}}{\mathrm{2}}.{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{{x}}{\mathrm{2}}{dx} \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)\mid{cos}\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{{x}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)\left\{{cos}\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}}−\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left\{\mathrm{1}−{cos}\left({n}\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$$$ \\ $$$${if}\:{n}={even} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:{or}\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${if}\:{n}\:{odd} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:\:{or}\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${now}\:{considering} \\ $$$${A}_{{n}} −{A}_{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${pls}\:{wait}…. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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