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Integration Questions

Question Number 120688 by bobhans last updated on 01/Nov/20

$$\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{x}\:\mathrm{dx}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\: \\$$$$\\$$

Answered by john santu last updated on 02/Nov/20

$$\:\psi\:=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{x}\:{dx}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}\:;\:\:\left[\:{let}\:{x}\:=\:\pi−{w}\:\right]\: \\$$$$\psi\:=\:\underset{\pi} {\overset{\mathrm{0}} {\int}}\:\frac{\left(\pi−{x}\right)\left(−{dw}\right)}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \left(\pi−{w}\right)^{\mathrm{2}} }\:=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\pi−{w}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {w}}\:{dw} \\$$$${adding}\:{the}\:{equation}\:{give} \\$$$$\mathrm{2}\psi\:=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\pi}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}\:{dx}\: \\$$$$\psi=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\pi}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}\:{dx}\:=\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{cosec}\:^{\mathrm{2}} {x}\:{dx}}{\mathrm{cosec}\:^{\mathrm{2}} {x}+\mathrm{1}} \\$$$$\psi=\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{cosec}\:^{\mathrm{2}} {x}\:{dx}}{\left(\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} {x}\right)+\mathrm{1}}\:=\:\frac{\pi}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\int}}\:\frac{\mathrm{1}\:{dt}}{\mathrm{2}+{t}^{\mathrm{2}} \:}\:;\:{where}\:{t}\:=\:\mathrm{cot}\:{x} \\$$$$\psi=\frac{\pi}{\mathrm{2}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)\:\mid_{−\infty} ^{\infty} \\$$$$\psi=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\left[\frac{\pi}{\mathrm{2}}−\left(−\frac{\pi}{\mathrm{2}}\right)\:\right]=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}}\:.\blacktriangle \\$$