What is the integral of sec³x?

Quick explanation

Write $\sec^3x$ as $\sec x\cdot\sec^2x$ . That gives us a factor whose antiderivative is $\tan x$ , while the remaining secant differentiates into $\sec x\tan x$ . The resulting integral contains $\int\sec^3x\,dx$ again—not a failure, but the useful trick.

Full derivation

Let $I=\int\sec^3x\,dx$ . Choose $u=\sec x$ and $dv=\sec^2x\,dx$ .

du=\sec x\tan x\,dx

v=\tan x

I=\sec x\tan x-\int\sec x\tan^2x\,dx

=\sec x\tan x-\int\sec x(\sec^2x-1)\,dx

=\sec x\tan x-I+\int\sec x\,dx

2I=\sec x\tan x+\ln\!\left|\sec x+\tan x\right|

\boxed{I=\frac12\sec x\tan x+\frac12\ln\!\left|\sec x+\tan x\right|+C}

Verification

Differentiate the result. The first term contributes $\frac12(\sec x\tan^2x+\sec^3x)$ . The logarithmic term differentiates to $\frac12\sec x$ . Using $\tan^2x+1=\sec^2x$ , everything combines to $\sec^3x$ .

Derivative of proposed answer

\frac12\sec x\tan^2x+\frac12\sec^3x+\frac12\sec x=\sec^3x

Verified

Common mistakes

Stopping when the integral returns.That is the moment to solve for $I$ , not abandon ship.
Forgetting the factor of $\frac12$ .You get $2I$ before the final division.
Dropping the absolute-value bars.The logarithmic term is $\ln\!\left|\sec x+\tan x\right|$ on its valid intervals.

Quick explanation

Full derivation

Verification

Common mistakes

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