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Question Number 116627 by mnjuly1970 last updated on 05/Oct/20

$$$$$$...nicecalculus...$$$$$$$$pleaseevaluate::$$$$$$$$\mathrm{\Phi }=\frac{{\left({\int }_0^{\frac{\pi }{2}}\frac{e^x+e^{-x}}{sin\left(x\right)+cos\left(x\right)}dx\right)}^2}{\left({\int }_0^{\frac{\pi }{2}}\frac{e^x}{sin\left(x\right)+cos\left(x\right)}dx\right)\left({\int }_0^{\frac{\pi }{2}}\frac{e^{-x}}{sin\left(x\right)+cos\left(x\right)}dx\right)}=???$$$$$$$$m.n.1970$$$$$$

Answered by mindispower last updated on 05/Oct/20

$${\int }_0^{\frac{\pi }{2}}\frac{e^x}{sin\left(x\right)+cos\left(x\right)}dx=J$$$$={\int }_0^{\frac{\pi }{2}}\frac{e^{\frac{\pi }{2}-x}}{cos\left(x\right)+sin\left(x\right)}dx$$$$=e^{\frac{\pi }{2}}{\int }_0^{\frac{\pi }{2}}\frac{e^{-x}}{sin\left(x\right)+cos\left(x\right)}=e^{\frac{\pi }{2}}I$$$$\mathrm{\Phi }=\frac{{(J+I)}^2}{I.J}=\frac{{({Ie}^{\frac{\pi }{2}}+I)}^2}{I.e^{\frac{\pi }{2}}I}=\frac{{(e^{\frac{\pi }{2}}+1)}^2}{e^{\frac{\pi }{2}}}$$$$=e^{\frac{\pi }{2}}+e^{-\frac{\pi }{2}}+2=2(ch\left(\frac{\pi }{2}\right)+1)$$

Commented by mnjuly1970 last updated on 05/Oct/20

$$voodforyou$$

Answered by Dwaipayan Shikari last updated on 05/Oct/20

$$I={\int }_0^{\frac{\pi }{2}}\frac{e^x+e^{-x}}{sinx+cosx}dx$$$$I_1={\int }_0^{\frac{\pi }{2}}\frac{e^x}{sinx+cosx}dx={\int }_0^{\frac{\pi }{2}}\frac{e^{\frac{\pi }{2}-x}}{sinx+cosx}dx=e^{\frac{\pi }{2}}{\int }_0^{\frac{\pi }{2}}\frac{e^{-x}}{sinx+cosx}dx$$$$I_2={\int }_0^{\frac{\pi }{2}}\frac{e^{-x}}{sinx+cosx}dx$$$$I_2=e^{\frac{\pi }{2}}{I}_1$$$$I_1+I_2=I\left(Byobservation\right)$$$$\mathrm{\Phi }=\frac{{\left(I\right)}^2}{I_1{I}_2}=\frac{{(I_1+I_2)}^2}{I_1^2{e}^{\frac{\pi }{2}}}=\frac{{(e^{\frac{\pi }{2}}+1)}^2}{e^{\frac{\pi }{2}}}=i^i.{(\sqrt[i]{i}+1)}^2$$

Commented by Dwaipayan Shikari last updated on 05/Oct/20

$${\mathrm{i}}^{\mathrm{i}}={\mathrm{e}}^{-\frac{\pi }{2}}and\sqrt[i]{i}=i^{\frac{1}{i}}=i^{-i}=e^{\frac{\pi }{2}}$$

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