## Calculus: Early Transcendentals 8th Edition

$$\int\frac{dx}{\cos x-1}=\csc x+\cot x+C$$
$$A=\int\frac{dx}{\cos x-1}$$ Multiply both numerator and denominator by $(\cos x+1)$ $$A=\int\frac{\cos x+1}{(\cos x-1)(\cos x+1)}dx$$ $$A=\int\frac{\cos x+1}{\cos^2 x-1}dx$$ Now we know that $\sin^2 x+\cos^2 x=1$. So, $\cos^2 x-1=-\sin^2 x$. That means $$A=-\int\frac{\cos x+1}{\sin^2 x}dx$$ $$A=-[\int\frac{\cos x}{\sin^2 x}dx+\int\frac{1}{\sin^2 x}dx]$$ $$A=-[X+Y]$$ *Consider X $$X=\int\frac{\cos x}{\sin^2 x}dx$$ Let $u=\sin x$. We would have $du=\cos xdx$. Therefore, $$X=\int\frac{1}{u^2}du$$ $$X=\int u^{-2}du$$ $$X=\frac{u^{-1}}{-1}+C$$ $$X=-\frac{1}{u}+C$$ $$X=-\frac{1}{\sin x}+C$$ *Consider Y $$Y=\int\frac{1}{\sin^2 x}dx$$ $$Y=\int\csc^2 xdx$$ $$Y=-\cot x+C$$ Combine X and Y, we have $$A=-[-\frac{1}{\sin x}-\cot x]+C$$ $$A=\frac{1}{\sin x}+\cot x+C$$ $$A=\csc x+\cot x+C$$