Indefinite integral of secant cubed

I need to calculate the following indefinite integral:

$$I=\int \frac{1}{\cos^3(x)}dx$$

I know what the result is (from Mathematica):

$$I=\tanh^{-1}(\tan(x/2))+(1/2)\sec(x)\tan(x)$$

but I don't know how to integrate it myself. I have been trying some substitutions to no avail.

Equivalently, I need to know how to compute:

$$I=\int \sqrt{1+z^2}dz$$

which follows after making the change of variables $z=\tan x$.

Solution 1:

We have an odd power of cosine. So there is a mechanical procedure for doing the integration. Multiply top and bottom by $\cos x$. The bottom is now $\cos^4 x$, which is $(1-\sin^2 x)^2$. So we want to find $$\int \frac{\cos x\,dx}{(1-\sin^2 x)^2}.$$ After the natural substitution $t=\sin x$, we arrive at $$\int \frac{dt}{(1-t^2)^2}.$$ So we want the integral of a rational function. Use the partial fractions machinery to find numbers $A$, $B$, $C$, $D$ such that $$\frac{1}{(1-t^2)^2}=\frac{A}{1-t}+\frac{B}{(1-t)^2}+ \frac{C}{1+t}+\frac{D}{(1+t)^2}$$ and integrate.

Solution 2:

Hint: rewrite the integral as

$$\int \sec ^3 (x) \, dx$$

Recall the identity $\sec^2(x)=\tan^2(x)+1$.

So, substituting, you get

$$\int\sec(x)(\tan^2(x)+1) \, dx=\int\tan(x)\tan(x)\sec(x) \, dx+\int\sec(x) \, dx.$$

The first integral can be solved by $u$-substitution and integration by parts, while the second, is an identity.

$$\int\tan(x) \, d\sec(x) = \tan(x)\sec(x)-\int\sec(x) \, d\tan(x)$$

But $\int\sec(x) \, d\tan(x)$ is the original integral. So write an equation and solve for $\int \sec^3(x)dx$

Solution 3:

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#c00000}{\int\sec^{3}\pars{x}\,\dd x}= \int\sec\pars{x}\,\dd\tan\pars{x} = \tan\pars{x}\sec\pars{x} - \int\tan\pars{x}\bracks{\sec\pars{x}\tan\pars{x}}\,\dd x \\[3mm]&= \tan\pars{x}\sec\pars{x} - \int\sec^{3}\pars{x}\,\dd x +\int\sec\pars{x}\,\dd x \\[3mm]&= \tan\pars{x}\sec\pars{x} - \color{#c00000}{\int\sec^{3}\pars{x}\,\dd x} + \ln\pars{\sec\pars{x} + \tan\pars{x}} \end{align}

$$\color{#0000ff}{\large% \int\sec^{3}\pars{x}\,\dd x = \half\bracks{\tan\pars{x}\sec\pars{x} + \ln\pars{\vphantom{\LARGE A}\sec\pars{x} + \tan\pars{x}}}} $$

Solution 4:

Immediate calculation of $$\int \sec ^{3}xdx.$$

We need the basic formulas of the first two derivatives of $\sec x:$\begin{eqnarray*} (\sec x)^{\prime } &=&\sec x\tan x \\ (\sec x)^{\prime \prime } &=&2\sec ^{3}x-\sec x \end{eqnarray*} Then \begin{eqnarray*} \int \sec ^{3}xdx &=&\frac{1}{2}\int \sec xdx+\frac{1}{2}\int (\sec x)^{\prime \prime }dx \\ &=&\frac{1}{2}\ln \left\vert \sec x+\tan x\right\vert +\frac{1}{2}(\sec x)^{\prime }+C \\ &=&\frac{1}{2}\ln \left\vert \sec x+\tan x\right\vert +\frac{1}{2}\sec x\tan x+C. \end{eqnarray*}

Solution 5:

It appears that Mathematica is using the "universal change" for trigonometric integrals $\tan(x/2)=t$: http://en.wikibooks.org/wiki/Calculus/Integration_techniques/Tangent_Half_Angle.