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Question Number 58346 by ANTARES VY last updated on 21/Apr/19

∫((cos(x))/(cos(2016°+cos(x)))×dx

$$\int\frac{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{cos}}\left(\mathrm{2016}°+\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\right.}×\boldsymbol{\mathrm{dx}} \\ $$

Answered by ANTARES VY last updated on 21/Apr/19

F(x)=?

$$\boldsymbol{\mathrm{F}}\left(\boldsymbol{\mathrm{x}}\right)=? \\ $$

Answered by tanmay last updated on 23/Apr/19

let cos(2016^o )=a  ∫((cosx)/(a+cosx))dx  ∫((a+cosx−a)/(a+cosx))dx  ∫dx−a∫(dx/(a+cosx))  ∫dx−a∫(dx/(a+((1−tan^2 (x/2))/(1+tan^2 (x/2)))))  ∫dx−a∫((sec^2 (x/2)dx)/(a+1+tan^2 (x/2)(a−1)))  pls look here a=cos(2016^o )<0  so  the rectification  will be as follows in red colour  condition a<0  ∫dx−(a/(a−1))∫((sec^2 (x/2)dx)/(a+1+tan^2 (x/2)×(a−1)))  ∫dx−(a/(a−1))∫((sec^2 (x/2))/(((a+1)/(a−1))+tan^2 (x/2)))dx  since a<0  so 1−a=+ve  ∫dx−(a/(a−1))∫((sec^2 (x/2))/(tan^2 (x/2)−(((a+1)/(1−a)))))  ∫dx+((2a)/(1−a))∫((d(tan(x/2)))/((tan(x/2))^2 −((√((a+1)/(1−a))) )^2 ))  x+((2a)/(1−a))×(1/(2(√((1+a)/(1−a)))))ln(((tan(x/2)−(√((1+a)/(1−a))))/(tan(x/2)+(√((1+a)/(1−a))))))+c  =x+(a/(√(1−a^2 )))ln(((tan(x/2)−(√((1+a)/(1−a))))/(tan(x/2)+(√((1−a)/(1−z))))))+c    put a=cos(11.2π)  x+((cos(11.2π))/(√(1−cos^2 (11.2)π)))ln(((tan(x/2)−(√((1+cos(11.2)π)/(1−cos(11.2)π))))/(tan(x/2)+(√((1−cos(11.2)π)/(1−cos(11.2)π))))))

$${let}\:{cos}\left(\mathrm{2016}^{{o}} \right)={a} \\ $$$$\int\frac{{cosx}}{{a}+{cosx}}{dx} \\ $$$$\int\frac{{a}+{cosx}−{a}}{{a}+{cosx}}{dx} \\ $$$$\int{dx}−{a}\int\frac{{dx}}{{a}+{cosx}} \\ $$$$\int{dx}−{a}\int\frac{{dx}}{{a}+\frac{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\int{dx}−{a}\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}}{{a}+\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\left({a}−\mathrm{1}\right)} \\ $$$${pls}\:{look}\:{here}\:{a}={cos}\left(\mathrm{2016}^{{o}} \right)<\mathrm{0} \\ $$$${so}\:\:{the}\:{rectification}\:\:{will}\:{be}\:{as}\:{follows}\:{in}\:{red}\:{colour} \\ $$$$\boldsymbol{{condition}}\:\boldsymbol{{a}}<\mathrm{0} \\ $$$$\int{dx}−\frac{{a}}{{a}−\mathrm{1}}\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}}{{a}+\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}×\left({a}−\mathrm{1}\right)} \\ $$$$\int{dx}−\frac{{a}}{{a}−\mathrm{1}}\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$${since}\:{a}<\mathrm{0}\:\:{so}\:\mathrm{1}−{a}=+{ve} \\ $$$$\int{dx}−\frac{{a}}{{a}−\mathrm{1}}\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\left(\frac{{a}+\mathrm{1}}{\mathrm{1}−{a}}\right)} \\ $$$$\int{dx}+\frac{\mathrm{2}{a}}{\mathrm{1}−{a}}\int\frac{{d}\left({tan}\frac{{x}}{\mathrm{2}}\right)}{\left({tan}\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\sqrt{\frac{{a}+\mathrm{1}}{\mathrm{1}−{a}}}\:\right)^{\mathrm{2}} } \\ $$$${x}+\frac{\mathrm{2}{a}}{\mathrm{1}−{a}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}}{ln}\left(\frac{{tan}\frac{{x}}{\mathrm{2}}−\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}}{{tan}\frac{{x}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}}\right)+{c} \\ $$$$={x}+\frac{{a}}{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{ln}\left(\frac{{tan}\frac{{x}}{\mathrm{2}}−\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}}{{tan}\frac{{x}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}−{a}}{\mathrm{1}−{z}}}}\right)+{c} \\ $$$$ \\ $$$${put}\:{a}={cos}\left(\mathrm{11}.\mathrm{2}\pi\right) \\ $$$${x}+\frac{{cos}\left(\mathrm{11}.\mathrm{2}\pi\right)}{\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} \left(\mathrm{11}.\mathrm{2}\right)\pi}}{ln}\left(\frac{{tan}\frac{{x}}{\mathrm{2}}−\sqrt{\frac{\mathrm{1}+{cos}\left(\mathrm{11}.\mathrm{2}\right)\pi}{\mathrm{1}−{cos}\left(\mathrm{11}.\mathrm{2}\right)\pi}}}{{tan}\frac{{x}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}−{cos}\left(\mathrm{11}.\mathrm{2}\right)\pi}{\mathrm{1}−{cos}\left(\mathrm{11}.\mathrm{2}\right)\pi}}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mr W last updated on 23/Apr/19

please check sir:  a=cos 2016° <1  a^2 −1<0  (√(a^2 −1))=????

$${please}\:{check}\:{sir}: \\ $$$${a}=\mathrm{cos}\:\mathrm{2016}°\:<\mathrm{1} \\ $$$${a}^{\mathrm{2}} −\mathrm{1}<\mathrm{0} \\ $$$$\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}=???? \\ $$

Commented by tanmay last updated on 23/Apr/19

i have not checked the value of cos(2016^o )  yes cos(2016^o )<0  so a<1   hence the rectification will be added in red colour

$${i}\:{have}\:{not}\:{checked}\:{the}\:{value}\:{of}\:{cos}\left(\mathrm{2016}^{{o}} \right) \\ $$$${yes}\:{cos}\left(\mathrm{2016}^{{o}} \right)<\mathrm{0} \\ $$$${so}\:{a}<\mathrm{1}\: \\ $$$${hence}\:{the}\:{rectification}\:{will}\:{be}\:{added}\:{in}\:{red}\:{colour} \\ $$$$ \\ $$

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