Integrating sec²x, sec x tan x and tan x

$\sec^{2}x$ integrates to tan x

By the quotient rule, $\tan x = \frac{{\sin x}}{{\cos x}}$ differentiates to $\frac{{\cos x \times \cos x - \sin x \times \left(-\sin x\right)}}{\cos^{2}x} = \frac{{\cos^{2}x + \sin^{2}x}}{\cos^{2}x} = \frac{1}{\cos^{2}x}$, which is $\sec^{2}x$.

Read backwards, $\int \sec^{2}x \,dx = \tan x + C$. The integral of $\sec^{2}x$ from 0 to $\frac{\pi}{4}$ is $\tan\left(\frac{\pi}{4}\right) - \tan 0 = 1 - 0 = 1$.

A multiple of x inside divides out as usual: $\int \sec^{2}\left(3x\right) \,dx = \frac{{\tan 3x}}{3} + C$. Check: $\frac{{\tan 3x}}{3}$ differentiates to $\frac{{3 \sec^{2}\left(3x\right)}}{3}$.

The gold curve is $y = \sec^{2}x$, shaded from x = 0 to $x = \frac{\pi}{4}$, where its height is 2. The plain curve is y = tan x, which starts at 0 and reaches 1 at $x = \frac{\pi}{4}$. That height, 1, is the shaded area.

sec x tan x integrates to sec x

$\sec x = \left(\cos x\right)^{-1}$, so by the chain rule it differentiates to $-\left(\cos x\right)^{-2} \times \left(-\sin x\right) = \frac{{\sin x}}{{\cos^{2}x}}$. That is $1 / \cos x$ times $\frac{{\sin x}}{{\cos x}}$, which is sec x tan x.

So $\int \sec x \tan x \,dx = \sec x + C$. The integral of sec x tan x from 0 to $\frac{\pi}{3}$ is $\sec\left(\frac{\pi}{3}\right) - \sec 0 = 2 - 1 = 1$.

The cosecant versions come the same way, each with a minus: cot x differentiates to $-\cosec^{2}x$ and cosec x to −cosec x cot x. So $\int \cosec^{2}x \,dx = -\cot x + C$ and $\int \cosec x \cot x \,dx = -\cosec x + C$.

tan x is a logarithm

tan x is not on the list of derivatives, but it is a fraction: $\tan x = \frac{{\sin x}}{{\cos x}}$. The denominator cos x differentiates to −sin x, so the numerator is minus the derivative of the denominator.

A derivative over its function integrates to a log, and the minus comes outside: $\int \tan x \,dx = -\ln|\cos x| + C$. Check: −ln|cos x| differentiates to $- \frac{{-\sin x}}{{\cos x}} = \tan x$.

The integral of tan x from 0 to $\frac{\pi}{3}$ is $-\ln\left(\cos\left(\frac{\pi}{3}\right)\right) + \ln\left(\cos 0\right) = -\ln \frac{1}{2} + 0 = \ln 2 \approx 0.693$.

$\cot x = \frac{{\cos x}}{{\sin x}}$ has the derivative of its denominator on top exactly, so there is no minus: $\int \cot x \,dx = \ln|\sin x| + C$.

The gold curve is y = tan x, shaded from x = 0 to $x = \frac{\pi}{3}$. The plain curve is y = −ln(cos x), which starts at 0; its height at $x = \frac{\pi}{3}$ is $\ln 2 \approx 0.693$, the shaded area.

Two answers that are one

Some books give $\int \tan x \,dx = \ln|\sec x| + C$ instead. It is the same function. A minus in front of a log turns the inside upside down, since $-\ln a = \ln\left(1 / a\right)$. So $-\ln|\cos x| = \ln|1 / \cos x|$, and $1 / \cos x = \sec x$.

$\tan^{2}x$ and sec x

$\tan^{2}x$ is not a derivative either, but the identity $1 + \tan^{2}x = \sec^{2}x$ turns it into one: $\tan^{2}x = \sec^{2}x - 1$, so $\int \tan^{2}x \,dx = \tan x - x + C$. From 0 to $\frac{\pi}{4}$ that is $1 - \frac{\pi}{4} \approx 0.215$.

sec x itself needs a trick. Multiply it by $\frac{{\sec x + \tan x}}{{\sec x + \tan x}}$, which is 1: the integrand becomes $\frac{{\sec^{2}x + \sec x \tan x}}{{\sec x + \tan x}}$. The numerator is exactly the derivative of the denominator, so $\int \sec x \,dx = \ln|\sec x + \tan x| + C$. From 0 to $\frac{\pi}{4}$ that is $\ln\left(\sqrt{2} + 1\right) - \ln 1 \approx 0.881$.

The usual mistakes

Giving the derivative instead of the integral. sec x tan x is what sec x differentiates to, so it cannot be the integral of $\sec^{2}x$.

Mixing up the family. $\sec^{2}x$ integrates to tan x and sec x tan x to sec x; differentiating each answer checks which is which.

Losing the minus on tan x. ln|cos x| differentiates to −tan x, so $\int \tan x \,dx = -\ln|\cos x| + C$.

Adding a minus to cot x. cos x is the derivative of sin x, sign and all, so $\int \cot x \,dx = \ln|\sin x| + C$.

Treating ln|sec x| and −ln|cos x| as different answers. They differ only in how they are written.