Question Number 200586 by Calculusboy last updated on 20/Nov/23
Answered by Frix last updated on 21/Nov/23
![∫_0 ^π (x/(2+tan^2 x))dx=∫_0 ^π xdx−∫_0 ^π (x/(1+cos^2 x)) I_1 =∫_0 ^π xdx=(π^2 /2) I_2 =∫_0 ^π (x/(1+cos^2 x))dx =^(t=2x) (1/2)∫_0 ^(2π) (t/(3+cos t))dt =^(u=2π−t) =(1/2)∫_(2π) ^0 ((u−2π)/(3+cos u))du=(1/2)∫_0 ^(2π) ((2π−u)/(3+cos u))du ⇒ [Renaming variables] I_2 =(1/2)((1/2)∫_0 ^(2π) (v/(3+cos v))dv+(1/2)∫_0 ^(2π) ((2π−v)/(3+cos v))dv)= =(π/2)∫_0 ^(2π) (dv/(3+cos v))=π∫_0 ^π (dv/(3+cos v)) =^(w=tan (v/2)) =π∫_0 ^∞ (dw/(w^2 +2))=(π/( (√2)))[tan^(−1) (w/( (√2)))]_0 ^∞ =(((√2)π^2 )/4) ⇒ ∫_0 ^π (x/(2+tan^2 x))dx=I_1 −I_2 =(((2−(√2))π^2 )/4)](https://www.tinkutara.com/question/Q200610.png)
$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}}{\mathrm{2}+\mathrm{tan}^{\mathrm{2}} \:{x}}{dx}=\underset{\mathrm{0}} {\overset{\pi} {\int}}{xdx}−\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$${I}_{\mathrm{1}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}{xdx}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}\:\overset{{t}=\mathrm{2}{x}} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{{t}}{\mathrm{3}+\mathrm{cos}\:{t}}{dt}\:\overset{{u}=\mathrm{2}\pi−{t}} {=} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{2}\pi} {\overset{\mathrm{0}} {\int}}\frac{{u}−\mathrm{2}\pi}{\mathrm{3}+\mathrm{cos}\:{u}}{du}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{\mathrm{2}\pi−{u}}{\mathrm{3}+\mathrm{cos}\:{u}}{du} \\ $$$$\Rightarrow \\ $$$$\left[\mathrm{Renaming}\:\mathrm{variables}\right] \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{{v}}{\mathrm{3}+\mathrm{cos}\:{v}}{dv}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{\mathrm{2}\pi−{v}}{\mathrm{3}+\mathrm{cos}\:{v}}{dv}\right)= \\ $$$$=\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{{dv}}{\mathrm{3}+\mathrm{cos}\:{v}}=\pi\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dv}}{\mathrm{3}+\mathrm{cos}\:{v}}\:\overset{{w}=\mathrm{tan}\:\frac{{v}}{\mathrm{2}}} {=}\: \\ $$$$=\pi\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dw}}{{w}^{\mathrm{2}} +\mathrm{2}}=\frac{\pi}{\:\sqrt{\mathrm{2}}}\left[\mathrm{tan}^{−\mathrm{1}} \:\frac{{w}}{\:\sqrt{\mathrm{2}}}\right]_{\mathrm{0}} ^{\infty} =\frac{\sqrt{\mathrm{2}}\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}}{\mathrm{2}+\mathrm{tan}^{\mathrm{2}} \:{x}}{dx}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} =\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$
Commented by Calculusboy last updated on 21/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Commented by Frix last updated on 21/Nov/23
��