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

\int udv=uv-\int vdu

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,

\int \sec^{3}{x}dx=\sec{x}\tan{x}-\int \sec^{3}{x}dx + \int\sec{x}dx

Grouping all integral terms with integrand sec^{3}{x} to the left gives us

\int \sec^{3}{x}dx+\sec^{3}{x}dx=\sec{x}\tan{x} + \int\sec{x}dx
2\int \sec^{3}{x}dx=\sec{x}\tan{x} + \int\sec{x}dx

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

2\int \sec^{3}{x}dx=\sec{x}\tan{x} + \ln|{sec{x}+\tan{x}}|

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:

\int 2x\sec^3{x^2}dx

Solution: Let u=x^2 such that du=2xdx. We then substitute u and du to the integral to get

\int 2x\sec^3{x^2}dx = \int \sec^3{u}du

Now we know the integral of secant cubed. Thus,

\int \sec^3{u}du
=\frac{1}{2}\sec{u}\tan{u} + \frac{1}{2}\ln|{sec{u}+\tan{u}}|

Putting back u=x^2 we get

\frac{1}{2}\sec{u}\tan{u} + \frac{1}{2}\ln|{sec{u}+\tan{u}}|
=\frac{1}{2}\sec{x^2}\tan{x^2} + \frac{1}{2}\ln|{sec{x^2}+\tan{x^2}}|

Thus,

\int 2x\sec^3{x^2}dx=\frac{1}{2}\sec{x^2}\tan{x^2} + \frac{1}{2}\ln|{sec{x^2}+\tan{x^2}}|

Example #2

Solve the following integral involving secant cubed.

\int \sec^3{\cos{y}}dx

Solution: Our secant function is dependent only on y while our variable of integration is x-dependent. Therefore,

\int \sec^3{\cos{y}}dx
= \sec^3{\cos{y}}\int dx
= x\sec^3{\cos{y}}

Example #3

Integrate.

\int -\sin{x}\sec^3{\cos{x}}dx

Solution: Let v=\cos{x} such that dv=-\sin{x}dx. Putting it to the integral, we get

\int -\sin{x}\sec^3{\cos{x}}dx = \int sec^3{v}dx

This is just an integral of a secant cubed for which the answer is given by

\int sec^3{v}dx =\frac{1}{2}\sec{v}\tan{v} + \frac{1}{2}\ln|{sec{v}+\tan{v}}|

We then put back v=\cos{x}, the final answer is

\int -\sin{x}\sec^3{(\cos{x})}dx=\frac{1}{2}\sec{(\cos{x})}\tan{(\cos{x})} + \frac{1}{2}\ln|{sec{(\cos{x})}+\tan{(\cos{x})}}|

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