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Evaluate \(\int \sec ^3 x d x\)
in Mathematics by Platinum (141,884 points) | 14 views

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Of course we already know the answer to this, but we needed to be clever to discover it. Here we'll use the new technique to discover the antiderivative. Let \(u=\sec x\) and \(d v=\sec ^2 x d x\). Then \(d u=\sec x \tan x d x\) and \(v=\tan x\) and
\[
\begin{aligned}
\int \sec ^3 x d x &=\sec x \tan x-\int \tan ^2 x \sec x d x \\
&=\sec x \tan x-\int\left(\sec ^2 x-1\right) \sec x d x \\
&=\sec x \tan x-\int \sec ^3 x d x+\int \sec x d x
\end{aligned}
\]

At first this looks useless - we're right back to \(\int \sec ^3 x d x\). But looking more closely:
\[
\begin{aligned}
\int \sec ^3 x d x &=\sec x \tan x-\int \sec ^3 x d x+\int \sec x d x \\
\int \sec ^3 x d x+\int \sec ^3 x d x &=\sec x \tan x+\int \sec x d x \\
2 \int \sec ^3 x d x &=\sec x \tan x+\int \sec x d x \\
\int \sec ^3 x d x &=\frac{\sec x \tan x}{2}+\frac{1}{2} \int \sec x d x \\
&=\frac{\sec x \tan x}{2}+\frac{\ln |\sec x+\tan x|}{2}+C .
\end{aligned}
\]
by Platinum (141,884 points)

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