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Question Number 52197 by maxmathsup by imad last updated on 04/Jan/19

$$calculate{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{cosx-sinx}{2+sin\left(2x\right)}dx$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19

$$2+sin2x$$$$1+{sin}^{2}x+{cos}^{2}x+2sinxcosx$$$$1+{(sinx+cosx)}^2$$$${\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{d(sinx+cosx)}{1+{(sinx+cosx)}^2}$$$$\mid {tan}^{-1}(sinx+cosx){\mid }_{\frac{\pi }{4}}^{\frac{\pi }{3}}$$$${tan}^{-1}(\frac{\sqrt{3}}{2}+\frac{1}{2})-{tan}^{-1}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})$$$${tan}^{-1}\left(\frac{1+\sqrt{3}}{2}\right)-{tan}^{-1}\left(\sqrt{2}\right)$$$${tan}^{-1}\left(\frac{\frac{1+\sqrt{3}}{2}-\sqrt{2}}{1+\frac{1+\sqrt{3}}{\sqrt{2}}}\right)$$$$={tan}^{-1}\left(\frac{\frac{1+\sqrt{3}-2\sqrt{2}}{2}}{\frac{\sqrt{2}+1+\sqrt{3}}{\sqrt{2}}}\right)$$$$={tan}^{-1}\left(\frac{1+\sqrt{3}-2\sqrt{2}}{\sqrt{2}(\sqrt{2}+1+\sqrt{3}}\right)$$$${tan}^{-1}\left(\frac{1+\sqrt{3}-2\sqrt{2}}{2+\sqrt{2}+\sqrt{6}}\right)$$

Commented by Abdo msup. last updated on 05/Jan/19

$$thankyousirTanmay.$$

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