Menu Close

0-pi-x-tan-2-x-1-dx-

Tinku Tara 0 Comments
Pin




Question Number 61408 by aliesam last updated on 02/Jun/19
∫_0 ^π (x/(tan^2 (x)−1)) dx
$$\int_{\mathrm{0}} ^{\pi} \frac{{x}}{{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{1}}\:{dx} \\ $$
Answered by tanmay last updated on 02/Jun/19
∫_0 ^π ((π−x)/(tan^2 (π−x)−1))dx =∫_0 ^π ((π−x)/(tan^2 (x)−1))dx 2I=∫_0 ^π (π/(tan^2 (x)−1))dx ((2I)/π)=∫_0 ^π (dx/(tan^2 (x)−1))=∫_0 ^π ((cos^2 x)/(sin^2 x−cos^2 x))dx ∫_0 ^(2a) f(x)dx=2∫_0 ^a f(x)dx [when f(2a−x)=f(x) ((2I)/π)=2∫_0 ^(π/2) ((cos^2 x)/(sin^2 x−cos^2 x))dx (I/π)=∫_0 ^(π/2) ((cos^2 ((π/2)−x))/(sin^2 ((π/2)−x)−cos^2 ((π/2)−x)))dx (I/π)=∫_0 ^(π/2) ((sin^2 x)/(cos^2 x−sin^2 x))dx=∫_0 ^(π/2) ((−sin^2 x)/(sin^2 x−cos^2 x))dx ((2I)/π)=∫_0 ^(π/2) ((cos^2 x−sin^2 x)/(sin^2 x−cos^2 x))dx ((2I)/π)=∣−x∣_0 ^(π/2) ((2I)/π)=−(π/2) I=((−π^2 )/4) pls check
$$\int_{\mathrm{0}} ^{\pi} \frac{\pi−{x}}{{tan}^{\mathrm{2}} \left(\pi−{x}\right)−\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{\pi−{x}}{{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{1}}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} \frac{\pi}{{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{1}}{dx} \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{1}}=\int_{\mathrm{0}} ^{\pi} \frac{{cos}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:\:\left[{when}\:{f}\left(\mathrm{2}{a}−{x}\right)={f}\left({x}\right)\right. \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\frac{{I}}{\pi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)−{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\ $$$$\frac{{I}}{\pi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{−{sin}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\mid−{x}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\frac{\mathrm{2}{I}}{\pi}=−\frac{\pi}{\mathrm{2}} \\ $$$${I}=\frac{−\pi^{\mathrm{2}} }{\mathrm{4}}\:{pls}\:{check} \\ $$$$ \\ $$
Commented by aliesam last updated on 02/Jun/19
thank you sir a brilliant sol
$${thank}\:{you}\:{sir}\:{a}\:{brilliant}\:{sol}\: \\ $$
Commented by tanmay last updated on 02/Jun/19
most welcome sir...
$${most}\:{welcome}\:{sir}… \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *