Question Number 201351 by sonukgindia last updated on 05/Dec/23
Answered by Sutrisno last updated on 05/Dec/23
$$=\int_{\mathrm{0}} ^{\pi} \frac{{x}}{\mathrm{1}+{sinx}}.\frac{\mathrm{1}−{sinx}}{\mathrm{1}−{sinx}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{x}−{xsinx}}{\mathrm{1}−{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{x}−{xsinx}}{{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} {xsec}^{\mathrm{2}} {x}−{xsinx}.{cos}^{−\mathrm{2}} {xdx}\:\left({use}\:{integral}\:{by}\:{parts}\right) \\ $$$$=\left({xtanx}−{ln}\mid{secx}\mid\right)−\left({xsecx}−{ln}\mid{secx}+{tanx}\right)\underset{\mathrm{0}} {\overset{\pi} {\mid}} \\ $$$$=\pi \\ $$
Answered by Calculusboy last updated on 05/Dec/23
![Solution: let y=𝛑−x dy=−dx when x=𝛑 y=0 and when x=0 y=𝛑 I=∫_𝛑 ^0 (((𝛑−y))/(1+sin(𝛑−y)))−dy ⇔ I=∫_0 ^𝛑 (((𝛑−y))/(1+sin(𝛑−y)))dy NB: sin(𝛑−y)=sin𝛑cosy−cos𝛑siny=siny I=∫_0 ^𝛑 (𝛑/(1+siny))dy−∫_0 ^𝛑 (y/(1+siny))dy I=𝛑∫_0 ^𝛑 (1/(1+siny))dy−I 2I=𝛑∫_0 ^𝛑 (1/(1+siny))dy (changing of variable) I=(𝛑/2)∫_0 ^𝛑 (1/(1+sinx))dx ⇔ I=(𝛑/2)∫_0 ^𝛑 (1/(1+sinx))∙((1−sinx)/(1−sinx))dx NB: (a+b)(a−b)=a^2 −b^2 ∴ (1+sinx)(1−sinx)=1−sin^2 x=cos^2 x I=(𝛑/2)∫_0 ^𝛑 ((1−sinx)/(1−sin^2 x))dx ⇔ I=(𝛑/2)∫_0 ^𝛑 ((1−sinx)/(cos^2 x))dx I=(𝛑/2)∫_0 ^𝛑 [(1/(cos^2 x))−((sinx)/(cos^2 x))]dx ⇔ I=(𝛑/2)∫_0 ^𝛑 [sec^2 x−secxtanx]dx I=(𝛑/2)[∫_0 ^𝛑 sec^2 xdx−∫_0 ^𝛑 secxtanxdx] I=(𝛑/2)[∣tanx∣_0 ^𝛑 −∣secx∣_0 ^𝛑 ] I=(𝛑/2){[tan(𝛑)−tan(0)]−[sec(𝛑)−sec(0)] I=(𝛑/2){(0−0)−(−1−1)} I=(𝛑/2){−(−2)} I=(𝛑/2)×2 I=𝛑](https://www.tinkutara.com/question/Q201364.png)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{{y}}=\boldsymbol{\pi}−\boldsymbol{{x}}\:\:\boldsymbol{{dy}}=−\boldsymbol{{dx}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\boldsymbol{\pi}\:\:\boldsymbol{{y}}=\mathrm{0}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\boldsymbol{{y}}=\boldsymbol{\pi} \\ $$$$\boldsymbol{{I}}=\int_{\boldsymbol{\pi}} ^{\mathrm{0}} \:\frac{\left(\boldsymbol{\pi}−\boldsymbol{{y}}\right)}{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{\pi}−\boldsymbol{{y}}\right)}−\boldsymbol{{dy}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \:\frac{\left(\boldsymbol{\pi}−\boldsymbol{{y}}\right)}{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{\pi}−\boldsymbol{{y}}\right)}\boldsymbol{{dy}} \\ $$$$\boldsymbol{{NB}}:\:\boldsymbol{{sin}}\left(\boldsymbol{\pi}−\boldsymbol{{y}}\right)=\boldsymbol{{sin}\pi{cosy}}−\boldsymbol{{cos}\pi{siny}}=\boldsymbol{{siny}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\boldsymbol{\pi}}{\mathrm{1}+\boldsymbol{{siny}}}\boldsymbol{{dy}}−\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \:\frac{\boldsymbol{{y}}}{\mathrm{1}+\boldsymbol{{siny}}}\boldsymbol{{dy}} \\ $$$$\boldsymbol{{I}}=\boldsymbol{\pi}\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{siny}}}\boldsymbol{{dy}}−\boldsymbol{{I}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\boldsymbol{\pi}\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{siny}}}\boldsymbol{{dy}}\:\:\left(\boldsymbol{{changing}}\:\boldsymbol{{of}}\:\boldsymbol{{variable}}\right) \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{sinx}}}\boldsymbol{{dx}}\:\:\:\Leftrightarrow\:\:\:\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{sinx}}}\centerdot\frac{\mathrm{1}−\boldsymbol{{sinx}}}{\mathrm{1}−\boldsymbol{{sinx}}}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{NB}}:\:\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)=\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \:\:\therefore\:\left(\mathrm{1}+\boldsymbol{{sinx}}\right)\left(\mathrm{1}−\boldsymbol{{sinx}}\right)=\mathrm{1}−\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}=\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}} \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \:\frac{\mathrm{1}−\boldsymbol{{sinx}}}{\mathrm{1}−\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}}\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \:\frac{\mathrm{1}−\boldsymbol{{sinx}}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \left[\frac{\mathrm{1}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}−\frac{\boldsymbol{{sinx}}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}\right]\boldsymbol{{dx}}\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \left[\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}−\boldsymbol{{secxtanx}}\right]\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\left[\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{xdx}}−\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \boldsymbol{{secxtanxdx}}\right] \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\left[\mid\boldsymbol{{tanx}}\mid_{\mathrm{0}} ^{\boldsymbol{\pi}} −\mid\boldsymbol{{secx}}\mid_{\mathrm{0}} ^{\boldsymbol{\pi}} \right] \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\left\{\left[\boldsymbol{{tan}}\left(\boldsymbol{\pi}\right)−\boldsymbol{{tan}}\left(\mathrm{0}\right)\right]−\left[\boldsymbol{{sec}}\left(\boldsymbol{\pi}\right)−\boldsymbol{{sec}}\left(\mathrm{0}\right)\right]\right. \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\left\{\left(\mathrm{0}−\mathrm{0}\right)−\left(−\mathrm{1}−\mathrm{1}\right)\right\} \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\left\{−\left(−\mathrm{2}\right)\right\} \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}×\mathrm{2} \\ $$$$\boldsymbol{{I}}=\boldsymbol{\pi} \\ $$$$ \\ $$