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Question Number 60685 by maxmathsup by imad last updated on 24/May/19
find I_n =∫_0 ^(π/2) ((1−cos(nx))/(sin^2 (nx)))dx
$${find}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}−{cos}\left({nx}\right)}{{sin}^{\mathrm{2}} \left({nx}\right)}{dx}\: \\ $$
Commented by maxmathsup by imad last updated on 25/May/19
we have I_n = ∫_0 ^(π/2) ((2sin^2 (((nx)/2)))/(4 sin^2 (((nx)/2))cos^2 (((nx)/2)))) dx =(1/2) ∫_0 ^(π/2) (dx/(cos^2 (((nx)/2)))) =_(((nx)/2)=t) (1/2) ∫_0 ^((nπ)/4) (1/(cos^2 t)) (2/n) dt = (1/n) ∫_0 ^((nπ)/4) (dt/((1+cos(2t))/2)) =(2/n) ∫_0 ^((nπ)/4) (dt/(1+cos(2t))) =_(tan(t) =u) (2/n) ∫_0 ^(tan(((nπ)/4))) (du/((1+u^2 )(1+((1−u^2 )/(1+u^2 ))))) =(2/n) ∫_0 ^(tan(((nπ)/4))) (du/(1+u^2 +1−u^2 )) =(1/n) ∫_0 ^(tan(((nπ)/4))) du =(1/n)tan(((nπ)/4)) ★ I_n =(1/n) tan(n(π/4)) ★ with n>0 .
$${we}\:{have}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{nx}}{\mathrm{2}}\right)}{\mathrm{4}\:{sin}^{\mathrm{2}} \left(\frac{{nx}}{\mathrm{2}}\right){cos}^{\mathrm{2}} \left(\frac{{nx}}{\mathrm{2}}\right)}\:{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dx}}{{cos}^{\mathrm{2}} \left(\frac{{nx}}{\mathrm{2}}\right)} \\ $$$$=_{\frac{{nx}}{\mathrm{2}}={t}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{{n}\pi}{\mathrm{4}}} \:\:\:\:\:\:\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {t}}\:\frac{\mathrm{2}}{{n}}\:{dt}\:=\:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\frac{{n}\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}\:=\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{\frac{{n}\pi}{\mathrm{4}}} \:\:\:\frac{{dt}}{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)} \\ $$$$=_{{tan}\left({t}\right)\:={u}} \:\:\:\:\:\:\:\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{{tan}\left(\frac{{n}\pi}{\mathrm{4}}\right)} \:\:\:\:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{{tan}\left(\frac{{n}\pi}{\mathrm{4}}\right)} \:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{{tan}\left(\frac{{n}\pi}{\mathrm{4}}\right)} \:{du}\:=\frac{\mathrm{1}}{{n}}{tan}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$$\bigstar\:{I}_{{n}} =\frac{\mathrm{1}}{{n}}\:{tan}\left({n}\frac{\pi}{\mathrm{4}}\right)\:\:\bigstar\:\:\:\:{with}\:{n}>\mathrm{0}\:. \\ $$
Answered by tanmay last updated on 24/May/19
∫_0 ^(π/2) ((2sin^2 ((nx)/2))/((2sin((nx)/2)cos((nx)/2))^2 ))dx (1/2)∫_0 ^(π/2) sec^2 ((nx)/2)dx (1/2)×∣tan((nx)/2)∣_0 ^(π/2) ×(2/n) (1/n)×tan((nπ)/4)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{nx}}{\mathrm{2}}}{\left(\mathrm{2}{sin}\frac{{nx}}{\mathrm{2}}{cos}\frac{{nx}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sec}^{\mathrm{2}} \frac{{nx}}{\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mid{tan}\frac{{nx}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} ×\frac{\mathrm{2}}{{n}} \\ $$$$\frac{\mathrm{1}}{{n}}×{tan}\frac{{n}\pi}{\mathrm{4}} \\ $$

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