Question-99168

Question Number 99168 by bemath last updated on 19/Jun/20

Answered by Kunal12588 last updated on 19/Jun/20

I=∫_0 ^( π/2) ((xcos x)/(1+sin^2 x))dx ⇒I=∫_0 ^( π/2) (((π/2)cos x)/(1+sin^2 x))dx−∫_0 ^( π/2) ((xcos x)/(1+sin^2 x))dx ⇒2I=(π/2)∫_0 ^( π/2) ((cos x)/(1+sin^2 x))dx ⇒I=−(π/4)∫_0 ^( π/2) ((d(sin x))/(1+sin^2 x)) ⇒I=−(π/4)tan^(−1) (sin (π/2))=−(π/4)tan^(−1) (1) ⇒I=−(π^2 /(16))

$${I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{{x}\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \:\frac{\left(\pi/\mathrm{2}\right)\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}−\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{{x}\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\Rightarrow\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\Rightarrow{I}=−\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{{d}\left(\mathrm{sin}\:{x}\right)}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}} \\ $$$$\Rightarrow{I}=−\frac{\pi}{\mathrm{4}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\right)=−\frac{\pi}{\mathrm{4}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\Rightarrow{I}=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$

Commented by mathmax by abdo last updated on 19/Jun/20

miss kunal if yiu have done the changement x=(π/2)−t we get I =∫_0 ^(π/2) ((((π/2)−t)sint)/(1+cos^2 t))dt =(π/2) ∫_0 ^(π/2) ((sint)/(1+cos^2 t))dt −∫_0 ^(π/2) ((tsint)/(1+cos^2 t)) dt how ∫_0 ^(π/2) ((xcosx)/(1+sin^2 x))dx =∫_0 ^(π/2) ((xsinx)/(1+cos^2 x))dx ?

$$\mathrm{miss}\:\mathrm{kunal}\:\mathrm{if}\:\mathrm{yiu}\:\mathrm{have}\:\mathrm{done}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\frac{\pi}{\mathrm{2}}−\mathrm{t}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)\mathrm{sint}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{t}}\mathrm{dt}\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{sint}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{t}}\mathrm{dt}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{tsint}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{t}}\:\mathrm{dt}\: \\ $$$$\mathrm{how}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{xcosx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{xsinx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:? \\ $$

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