The Integral of Secant (Four Derivations)

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On this page, I am providing four different methods to show the following integral result.

$$\int \sec xdx=\boxed{\ln |\sec x+\tan x|+C}=\boxed{\ln \left|\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right|+C}$$

This integral is particularly interesting because despite its simple appearance, there are a variety of different methods available to evaluate it, though none of the methods are all that easy to work out initially.

Method 1: the conventional trick

Multiply first by the factor $\frac{\sec x+\tan x}{\sec x+\tan x}$.

$$\int \sec xdx$$ $$\int \sec x\cdot \frac{\sec x+\tan x}{\sec x+\tan x}dx$$ $$\int \frac{\sec x\tan x+{\sec }^{2}x}{\sec x+\tan x}dx$$

Now, notice that the numerator is the derivative of the denominator, so we can use the pattern $\int \frac{u^{\prime }}{u}du=\ln |u|+C.$

$\boxed{\ln |\sec x+\tan x|+C}$

Method 2: substitution $u=\sec x$

Let $u=\sec x$, so $du=\sec x\tan xdx,$ which gives $\frac{1}{\tan x}dx=\sec xdx$. Since $\tan x=\sqrt{{\sec }^{2}x-1}$, we can hence write $\frac{1}{\sqrt{u^2-1}}du=\sec xdx$, which can be put into the integral.

$$\int \sec xdx=\int \frac{1}{\sqrt{u^2-1}}du$$

This is a well-known integral.

$$\text{arccosh}u+C$$

Switch to the logarithmic version of $\text{arccosh}$.

$$\ln \left|u+\sqrt{u^2-1}\right|+C$$

Back-substitute $u=\sec x$, taking advantage of the identity $\sqrt{{\sec }^{2}x-1}=\tan x$.

$$\boxed{\ln |\sec x+\tan x|+C}$$

Method 3: substitution $u=\tan x$

Let $u=\tan x$, so $du={\sec }^{2}xdx$ and hence $\frac{1}{\sec x}du=\sec xdx$. Since $\sec x=\sqrt{1+{\tan }^{2}x}$, this can be rewritten as $\frac{1}{\sqrt{1+u^2}}du=\sec xdx$ and put into the integral.

$$\int \sec xdx=\int \frac{1}{\sqrt{1+u^2}}du$$

This is a well-known integral.

$$\text{arcsinh}u+C$$

Switch to the logarithmic version of $\text{arcsinh}$.

$$\ln \left|\sqrt{1+u^2}+u\right|+C$$

Back-substitute $u=\tan x$, taking advantage of the identity $\sqrt{1+{\tan }^{2}x}=\sec x$.

$$\boxed{\ln \left|\sec x+\tan x\right|+C}$$

Method 4: via Weierstrass substitution $u=\tan \frac{x}{2}$

Use the double-angle identity to expand $\sec x$ in terms of $\sec \frac{x}{2}$ and $\tan \frac{x}{2}$.

$$\sec x=\frac{1}{\cos x}=\frac{1}{\cos (2\cdot \frac{x}{2})}=\frac{1}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}\cdot \frac{\frac{1}{{\cos }^2\frac{x}{2}}}{\frac{1}{{\cos }^2\frac{x}{2}}}=\frac{{\sec }^2\frac{x}{2}}{1-{\tan }^2\frac{x}{2}}$$

Hence,

$$\int \sec xdx=\int \frac{{\sec }^2\frac{x}{2}}{1-{\tan }^2\frac{x}{2}}dx.$$

Let $u=\tan \frac{x}{2}$, so $2du={\sec }^2\frac{x}{2}dx$.

$$\int \frac{2}{1-u^2}du$$

Perform a partial fraction decomposition and integrate.

$$\int \frac{1}{1+u}+\frac{1}{1-u}du$$ $$\ln \left|\frac{1+u}{1-u}\right|+C$$

Back-substitute $u=\tan \frac{x}{2}$.

$$\boxed{\ln \left|\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right|+C}$$

This result is equivalent to $\ln |\sec x+\tan x|+C$. Click below to see a proof.

Proof of equivalence to $\ln \left|\sec x+\tan x\right|+C$

$$\ln \left|\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\cdot \frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right|+C$$ $$\ln \left|\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\cdot \frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right|+C$$ $$\ln \left|\frac{(\cos \frac{x}{2}+\sin \frac{x}{2}{)}^2}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}\right|+C$$ $$\ln \left|\frac{1+2\cos \frac{x}{2}\sin \frac{x}{2}}{\cos x}\right|+C$$ $$\ln \left|\frac{1+\sin x}{\cos x}\right|+C$$ $$\ln \left|\sec x+\tan x\right|+C$$