Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 76
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Find the integral $\displaystyle \int \frac{\sin x}{1 + \cos ^2 x} dx$
If we let $u = \cos x $, then $\displaystyle du = - \sin x dx$, so $\sin x dx = -du$. Thus,
$ \begin{equation} \begin{aligned} \int \frac{\sin x}{1 + \cos^2 x} dx =& \int \frac{1}{1 + \cos^2 x} \sin x dx \\ \\ \int \frac{\sin x}{1 + \cos^2 x} dx =& \int \frac{1}{1 + u^2} \cdot -du \\ \\ \int \frac{\sin x}{1 + \cos^2 x} dx =& - \int \frac{1}{1 + u^2} du \\ \\ \int \frac{\sin x}{1 + \cos^2 x} dx =& - \tan^{-1} u + C \\ \\ \int \frac{\sin x}{1 + \cos^2 x} dx =& - \tan^{-1} (\cos x) + C \end{aligned} \end{equation} $
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