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Question Number 52197 by maxmathsup by imad last updated on 04/Jan/19
calculate ∫_(π/4) ^(π/3)  ((cosx −sinx)/(2 +sin(2x)))dx
$${calculate}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{cosx}\:−{sinx}}{\mathrm{2}\:+{sin}\left(\mathrm{2}{x}\right)}{dx} \\ $$$$ \\ $$
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   ∫_(π/4) ^(π/3)  ((d(sinx+cosx))/(1+(sinx+cosx)^2 ))  ∣tan^(−1) (sinx+cosx)∣_(π/4) ^(π/3)   tan^(−1) (((√3)/2)+(1/2))−tan^(−1) ((1/( (√2)))+(1/( (√2))))  tan^(−1) (((1+(√3))/2))−tan^(−1) ((√2) )  tan^(−1) (((((1+(√3))/2)−(√2))/(1+((1+(√3))/( (√2))))))  =tan^(−1) ((((1+(√3) −2(√2))/2)/(((√2) +1+(√3))/( (√2)))))  =tan^(−1) (((1+(√3) −2(√2))/( (√2) ((√2) +1+(√3))))  tan^(−1) (((1+(√3) −2(√2))/(2+(√2) +(√6))))
$$\mathrm{2}+{sin}\mathrm{2}{x} \\ $$$$\mathrm{1}+{sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}+\mathrm{2}{sinxcosx} \\ $$$$\mathrm{1}+\left({sinx}+{cosx}\right)^{\mathrm{2}} \\ $$$$\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{d}\left({sinx}+{cosx}\right)}{\mathrm{1}+\left({sinx}+{cosx}\right)^{\mathrm{2}} } \\ $$$$\mid{tan}^{−\mathrm{1}} \left({sinx}+{cosx}\right)\mid_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\:\right) \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}−\sqrt{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}+\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{2}}\:+\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}\:\left(\sqrt{\mathrm{2}}\:+\mathrm{1}+\sqrt{\mathrm{3}}\right.}\right) \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{6}}}\right) \\ $$
Commented by Abdo msup. last updated on 05/Jan/19
thank you sir Tanmay.
$${thank}\:{you}\:{sir}\:{Tanmay}. \\ $$

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