Integration Questions

Question Number 106932 by Ar Brandon last updated on 07/Aug/20

$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\$$

Answered by mathmax by abdo last updated on 08/Aug/20

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{xdx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\:\mathrm{changement}\:\mathrm{x}\:=\pi−\mathrm{t}\:\mathrm{give} \\$$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\left(\pi−\mathrm{t}\right)\mathrm{dt}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{t}}\:=\pi\:\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{t}}−\mathrm{I}\:\Rightarrow\mathrm{2I}\:=\pi\:\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{t}} \\$$$$\mathrm{but}\:\mathrm{we}\:\mathrm{have}\:\mathrm{proved}\:\:\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{t}}\:=\frac{\pi}{\sqrt{\mathrm{2}}}\:\Rightarrow\mathrm{2I}\:=\frac{\pi^{\mathrm{2}} }{\sqrt{\mathrm{2}}}\:\Rightarrow\mathrm{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}} \\$$

Answered by Dwaipayan Shikari last updated on 08/Aug/20

$$\int_{\mathrm{0}} ^{\pi} \frac{{x}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}=\int\frac{\pi−{x}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}={I} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} \frac{\pi}{\mathrm{1}+{cos}^{\mathrm{2}} {x}} \\$$$$\mathrm{2}{I}=\pi\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {x}}{{sec}^{\mathrm{2}} {x}+\mathrm{1}} \\$$$$\mathrm{2}{I}=\pi\int_{\mathrm{0}} ^{\pi} \frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\left({tanx}={t}\right. \\$$$$\mathrm{2}{I}=\frac{\pi}{\sqrt{\mathrm{2}}}\left[{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tanx}}{\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\pi} =\frac{\pi}{\sqrt{\mathrm{2}}}.\pi=\frac{\pi^{\mathrm{2}} }{\sqrt{\mathrm{2}}} \\$$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}} \\$$

Commented by Ar Brandon last updated on 08/Aug/20

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Answered by Ar Brandon last updated on 08/Aug/20

$$\mathcal{J}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\$$$$\:\:\:\:=\int_{\mathrm{0}} ^{\pi} \frac{\pi−\mathrm{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}=\int_{\mathrm{0}} ^{\pi} \frac{\pi}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}−\mathcal{J} \\$$$$\mathrm{2}\mathcal{J}=\int_{\mathrm{0}} ^{\pi} \frac{\pi}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}=\pi\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\mathrm{sec}^{\mathrm{2}} \mathrm{x}+\mathrm{1}}\mathrm{dx} \\$$$$\:\:\:\:\:\:\:=\pi\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\mathrm{2}}\mathrm{dx}=\pi\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{d}\left(\mathrm{tanx}\right)}{\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\mathrm{2}}\mathrm{dx} \\$$$$\:\:\:\:\:\:\:=\pi\left[\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{Arctan}\left(\frac{\mathrm{tanx}}{\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\pi} \\$$$$\:\:\:\:\:\:\:=\frac{\pi}{\sqrt{\mathrm{2}}}\left[\mathrm{Arctan}\left(−\frac{\mathrm{0}}{\sqrt{\mathrm{2}}}\right)−\mathrm{Arctan}\left(\frac{\mathrm{0}}{\sqrt{\mathrm{2}}}\:\right)\right]=\frac{\pi\left(\pi−\mathrm{0}\right)}{\sqrt{\mathrm{2}}}=\frac{\pi^{\mathrm{2}} }{\sqrt{\mathrm{2}}} \\$$$$\mathcal{J}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}} \\$$