Question Number 201290 by sonukgindia last updated on 03/Dec/23
Answered by Calculusboy last updated on 03/Dec/23
$$\boldsymbol{{Solution}}:\:\boldsymbol{{By}}\:\boldsymbol{{using}}\:\boldsymbol{{kings}}\:\boldsymbol{{rule}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)+\boldsymbol{{cos}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}\boldsymbol{{dx}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\boldsymbol{{cos}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}{\boldsymbol{{cos}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)+\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)+\boldsymbol{{cos}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)+\boldsymbol{{cos}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{1}\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\left(\boldsymbol{{x}}\right)\mid_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} +\boldsymbol{{C}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\mathrm{0}\right)\:\:\:\:\Leftrightarrow\:\:\:\mathrm{2}\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$$$ \\ $$