Question Number 43418 by Raj Singh last updated on 10/Sep/18
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
![∫((cosxdx)/(sinxcosx+1)) ∫((2cosx)/(2cosxsinx+2))dx ∫((cosx+sinx+cosx−sinx)/(1+(sinx+cosx)^2 ))dx ∫((cosx−sinx)/(1+(sinx+cosx)^2 ))dx+∫((cosx+sinx)/(2+2sinxcosx))dx ∫((d(sinx+cosx))/(1+(sinx+cosx)))−∫(((cosx+sinx)dx)/(−3+1−2sinxcosx)) ∫((d(sinx+cosx))/(1+(sinx+cosx)^2 ))−∫((d(sinx−cosx))/(−3+(sinx−cosx)^2 )) formula ∫(dx/(x^2 +a^2 ))=(1/a)tan^− ((x/a)) ∫(dx/(x^2 −b^2 )) =(1/(2b))∫(((x+b)−(x−b))/((x+b)(x−b)))dx =(1/(2b))[∫(dx/(x−b))−∫(dx/(x+b))] =(1/(2b))ln∣((x−b)/(x+b))∣ so ans is tan^(−1) (sinx+cosx)−(1/(2(√3) ))ln∣((sinx−cosx−(√3) )/(sinx−cosx+(√3)))∣+c](https://www.tinkutara.com/question/Q43437.png)
$$\int\frac{{cosxdx}}{{sinxcosx}+\mathrm{1}} \\ $$$$\int\frac{\mathrm{2}{cosx}}{\mathrm{2}{cosxsinx}+\mathrm{2}}{dx} \\ $$$$\int\frac{{cosx}+{sinx}+{cosx}−{sinx}}{\mathrm{1}+\left({sinx}+{cosx}\right)^{\mathrm{2}} }{dx} \\ $$$$\int\frac{{cosx}−{sinx}}{\mathrm{1}+\left({sinx}+{cosx}\right)^{\mathrm{2}} }{dx}+\int\frac{{cosx}+{sinx}}{\mathrm{2}+\mathrm{2}{sinxcosx}}{dx} \\ $$$$\int\frac{{d}\left({sinx}+{cosx}\right)}{\mathrm{1}+\left({sinx}+{cosx}\right)}−\int\frac{\left({cosx}+{sinx}\right){dx}}{−\mathrm{3}+\mathrm{1}−\mathrm{2}{sinxcosx}} \\ $$$$\int\frac{{d}\left({sinx}+{cosx}\right)}{\mathrm{1}+\left({sinx}+{cosx}\right)^{\mathrm{2}} }−\int\frac{{d}\left({sinx}−{cosx}\right)}{−\mathrm{3}+\left({sinx}−{cosx}\right)^{\mathrm{2}} } \\ $$$${formula}\:\int\frac{{dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}}{tan}^{−} \left(\frac{{x}}{{a}}\right) \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{b}}\int\frac{\left({x}+{b}\right)−\left({x}−{b}\right)}{\left({x}+{b}\right)\left({x}−{b}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{b}}\left[\int\frac{{dx}}{{x}−{b}}−\int\frac{{dx}}{{x}+{b}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{b}}{ln}\mid\frac{{x}−{b}}{{x}+{b}}\mid \\ $$$${so}\:{ans}\:{is} \\ $$$${tan}^{−\mathrm{1}} \left({sinx}+{cosx}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\:}{ln}\mid\frac{{sinx}−{cosx}−\sqrt{\mathrm{3}}\:}{{sinx}−{cosx}+\sqrt{\mathrm{3}}}\mid+{c} \\ $$$$ \\ $$