Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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)

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{nx}}{\mathrm{2}}}{\left(\mathrm{2}{sin}\frac{{nx}}{\mathrm{2}}{cos}\frac{{nx}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sec}^{\mathrm{2}} \frac{{nx}}{\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mid{tan}\frac{{nx}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} ×\frac{\mathrm{2}}{{n}} \\ $$$$\frac{\mathrm{1}}{{n}}×{tan}\frac{{n}\pi}{\mathrm{4}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com