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Question Number 152291 by iloveisrael last updated on 27/Aug/21

$$\int \frac{dx}{\cos x+\mathrm{cosec}x}=?$$

Answered by puissant last updated on 27/Aug/21

$$I=\int \frac{dx}{cosx+cosecx}$$$$=\int \frac{sinx}{cosxsinx+1}dx=\int \frac{2sinx}{sin2x+2}dx$$$$=\int \frac{sinx+cosx}{sin2x+2}dx+\int \frac{sinx-cosx}{sin2x+2}dx$$$$t=sinx-cosx\to dt=(sinx+cosx)dx$$$$t^2=1-sin2x\Rightarrow sin2x=1-t^2$$$$u=sinx+cosx\to -du=(sinx-cosx)dx$$$$u^2=1+sin2x\to sin2x=u^2-1$$$$$$$$\Rightarrow I=\int \frac{dt}{1-t^2+2}-\int \frac{du}{u^2-1+2}$$$$=\int \frac{dt}{{\left(\sqrt{3}\right)}^2-{\left(t\right)}^2}-\int \frac{du}{{\left(u\right)}^2+{\left(1\right)}^2}$$$$=\frac{1}{2\sqrt{3}}ln\mid \frac{\sqrt{3}-t}{\sqrt{3}+t}\mid -arctan\left(u\right)+C$$$$$$$$\therefore \because I=\frac{1}{2\sqrt{3}}ln\mid \frac{\sqrt{3}-(sinx-cosx)}{\sqrt{3}+(sinx-cosx)}\mid -arctan(sinx+cosx)+C..$$

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