Integration Questions

Question Number 61408 by aliesam last updated on 02/Jun/19

$$\int_{\mathrm{0}} ^{\pi} \frac{{x}}{{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{1}}\:{dx} \\$$

Answered by tanmay last updated on 02/Jun/19

$$\int_{\mathrm{0}} ^{\pi} \frac{\pi−{x}}{{tan}^{\mathrm{2}} \left(\pi−{x}\right)−\mathrm{1}}{dx} \\$$ $$=\int_{\mathrm{0}} ^{\pi} \frac{\pi−{x}}{{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{1}}{dx} \\$$ $$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} \frac{\pi}{{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{1}}{dx} \\$$ $$\frac{\mathrm{2}{I}}{\pi}=\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{1}}=\int_{\mathrm{0}} ^{\pi} \frac{{cos}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}{dx} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:\:\left[{when}\:{f}\left(\mathrm{2}{a}−{x}\right)={f}\left({x}\right)\right. \\$$ $$\frac{\mathrm{2}{I}}{\pi}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}{dx} \\$$ $$\frac{{I}}{\pi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)−{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\$$ $$\frac{{I}}{\pi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{−{sin}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}{dx} \\$$ $$\frac{\mathrm{2}{I}}{\pi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}{dx} \\$$ $$\frac{\mathrm{2}{I}}{\pi}=\mid−{x}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$ $$\frac{\mathrm{2}{I}}{\pi}=−\frac{\pi}{\mathrm{2}} \\$$ $${I}=\frac{−\pi^{\mathrm{2}} }{\mathrm{4}}\:{pls}\:{check} \\$$ $$\\$$

Commented byaliesam last updated on 02/Jun/19

$${thank}\:{you}\:{sir}\:{a}\:{brilliant}\:{sol}\: \\$$

Commented bytanmay last updated on 02/Jun/19

$${most}\:{welcome}\:{sir}... \\$$