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Evaluate $\int \sec ^3 x d x$
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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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