Secant-cubed function

From Calculus
(Redirected from Sec^3)

This article is about a particular function from a subset of the real numbers to the real numbers. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article.
View a complete list of particular functions on this wiki

For functions involving angles (trigonometric functions, inverse trigonometric functions, etc.) we follow the convention that all angles are measured in radians. Thus, for instance, the angle of

90

is measured as

π/2

.

Definition

The function, denoted sec3, is defined as the composite of the cube function and the secant function (which in turn is the composite of the reciprocal function and the cosine function). Explicitly, it is given as:

x(secx)3=1(cosx)3

Differentiation

First derivative

We can differentiate this function as a composite function, using the chain rule for differentiation. We get:

ddx(sec3x)=[3(secx)2](secxtanx)=3sec3xtanx

Alternatively, we could consider it as 1/(cosx)3, and hence as a composite of t1/t3 and cos. With that interpretation, the chain rule for differentiation yields:

3cos4x(sinx)=3sinxcos4x

The two answers can be verified to be equivalent.

Integration

First antiderivative

We use the recursive version of integration by parts.

We rewrite sec3x=secxsec2x and perform integration by parts, taking sec2 as the part to integrate. We use that an antiderivative of sec2x is tanx whereas the derivative of secx is secxtanx:

sec3xdx=secx(sec2x)dx=secxtanx(secxtanx)tanxdx=secxtanxsecxtan2xdx

We now use the fact that tan2x+1=sec2x, or more explicitly, tan2x=sec2x1, to rewrite this as:

sec3xdx=secxtanxsecx(sec2x1)dx=secxtanxxsec3xdx+secxd

We now use the integration of the secant function to simplify this as:

sec3xdx=secxtanxsec3xdx+ln|secx+tanx|

We can choose an antiderivative I of sec3 so that the above equality (between the left-most and right-most expression) holds without any additive constant adjustment, and we get:

I=secxtanxI+ln|secx+tanx|

We rearrange and obtain:

2I=secxtanx+ln|secx+tanx|

Dividing by 2, we get:

I=secxtanx+ln|secx+tanx|2

The general antiderivative expression is thus:

secxtanx+ln|secx+tanx|2+C