Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 96

Evaluate $\displaystyle \int^{\frac{\pi}{2}}_0 \frac{\cos x}{1 + \sin^2 x} dx$


$
\begin{equation}
\begin{aligned}

\text{If we let } u =& \sin x, \text{ then}
\\
\\
du =& \cos x dx

\end{aligned}
\end{equation}
$


Make sure that the upper and lower limits are also in terms of $u$.

So,


$
\begin{equation}
\begin{aligned}

\int^{\frac{\pi}{2}}_0 \frac{\cos x}{1 + \sin^2 x} =& \int^{\sin \left( \frac{\pi}{2} \right)}_{\sin (0)} \frac{du}{1 + u^2}
\\
\\
=& \int^1_0 \frac{du}{1 + u^2}

\end{aligned}
\end{equation}
$



Recall that


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (\tan^{-1} (x)) =& \frac{1}{1 + x^2}
\\
\\
=& \left[ \tan^{-1} u \right]^1_0
\\
\\
=& \tan^{-1} (1) - \tan^{-1} (0)
\\
\\
=& \frac{\pi}{4}

\end{aligned}
\end{equation}
$