Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 76

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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