Subject: Calculus
Integral Of Secant Cubed
The integral of secant cubed is given below.
\int \sec^{3}{x}dx=\frac{1}{2}\sec{x}\tan{x}+\frac{1}{2}\ln\left | \sec{x}+\tan{x} \right |
In this article, let me show the details of the solution of the above integral which can be done via many integration methods. But in this case, let me show you how its done via integration by parts.
\int \sec^{3}{x}dx
Let u=\sec{x} such that du=\sec{x}\tan{x}dx and dv=sec^2{x} such that v=\tan{x}
then according to the integration by parts, such integral is expressed as
Then, we can proceed as \int \sec^{3}{x}dx is given by

Plot of a secant cubed function.
uv-\int vdu
=\sec{x}\tan{x}-\int \sec{x}\tan^2{x}dx
=\sec{x}\tan{x}-\int \sec{x}(\sec^2{x}-1)dx
;since \tan^2{x}=\sec^2{x}-1, then
=\sec{x}\tan{x}-\left (\int \sec^{3}{x}dx - \int\sec{x}dx\right )
=\sec{x}\tan{x}-\int \sec^{3}{x}dx + \int\sec{x}dx
So the whole picture of the integral looks like,
Grouping all integral terms with integrand sec^{3}{x} to the left gives us
Then we perform the integral operation in the right side. Since we know that \int sec{x}dx=\ln|{sec{x}+\tan{x}}|, the integral above would be
Dividing by two, we get
\int \sec^{3}{x}dx=\frac{1}{2}\sec{x}\tan{x} + \frac{1}{2}\ln|{sec{x}+\tan{x}}|
Which is just what we want to show!.
Example #1
Find the integral of the following:
Solution: Let u=x^2 such that du=2xdx. We then substitute u and du to the integral to get
Now we know the integral of secant cubed. Thus,
Putting back u=x^2 we get
Thus,
Example #2
Solve the following integral involving secant cubed.
Solution: Our secant function is dependent only on y while our variable of integration is x-dependent. Therefore,
Example #3
Integrate.
Solution: Let v=\cos{x} such that dv=-\sin{x}dx. Putting it to the integral, we get
This is just an integral of a secant cubed for which the answer is given by
We then put back v=\cos{x}, the final answer is
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