## Calculus: Early Transcendentals 8th Edition

$$\int\frac{\sin x}{1+\cos^2 x}dx=\cot^{-1}(\cos x)+C$$
$$A=\int\frac{\sin x}{1+\cos^2 x}dx$$ Let $u=\cos x$ Then we have $du=-\sin x'dx$. That means $\sin x dx=-du$ Substitute into $A$: $$A=\int-\frac{1}{1+u^2}du$$ The differentiation of $\cot^{-1}x$ is $-\frac{1}{1+x^2}$. Therefore, since $A=\int-\frac{1}{1+u^2}du$, $$A=\cot^{-1}u+C$$ $$A=\cot^{-1}(\cos x)+C$$