Integrating sec²x, sec x tan x and tan x

${\sec }^{2}x$sec²x integrates to tan x

By the quotient rule, $\tan x=\frac{\sin x}{\cos x}$tan x = (sin x)/(cos x) differentiates to $\frac{\cos x\times \cos x-\sin x\times \left(-\sin x\right)}{{\cos }^{2}x}=\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}$(cos x × cos x − sin x × (−sin x))/cos²x = (cos²x + sin²x)/cos²x = 1/cos²x, which is ${\sec }^{2}x$sec²x.

Read backwards, $\int {\sec }^{2}xdx=\tan x+C$∫ sec²x dx = tan x + C. The integral of ${\sec }^{2}x$sec²x from 0 to $\frac{\pi }{4}$π/4 is $\tan \left(\frac{\pi }{4}\right)-\tan 0=1-0=1$tan(π/4) − tan 0 = 1 − 0 = 1.

A multiple of x inside divides out as usual: $\int {\sec }^2\left(3x\right)dx=\frac{\tan 3x}{3}+C$∫ sec²(3x) dx = (tan 3x)/3 + C. Check: $\frac{\tan 3x}{3}$(tan 3x)/3 differentiates to $\frac{3{\sec }^2\left(3x\right)}{3}$(3 sec²(3x))/3.

The gold curve is $y={\sec }^{2}x$y = sec²x, shaded from x = 0 to $x=\frac{\pi }{4}$x = π/4, where its height is 2. The plain curve is y = tan x, which starts at 0 and reaches 1 at $x=\frac{\pi }{4}$x = π/4. That height, 1, is the shaded area.

sec x tan x integrates to sec x

$\sec x={\left(\cos x\right)}^{-1}$sec x = (cos x)⁻¹, so by the chain rule it differentiates to $-{\left(\cos x\right)}^{-2}\times \left(-\sin x\right)=\frac{\sin x}{{\cos }^{2}x}$−(cos x)⁻² × (−sin x) = (sin x)/(cos²x). That is $1/\cos x$1 / cos x times $\frac{\sin x}{\cos x}$(sin x)/(cos x), which is sec x tan x.

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

The cosecant versions come the same way, each with a minus: cot x differentiates to $-{\mathrm{cosec}}^{2}x$−cosec²x and cosec x to −cosec x cot x. So $\int {\mathrm{cosec}}^{2}xdx=-\cot x+C$∫ cosec²x dx = −cot x + C and $\int \mathrm{cosec}x\cot xdx=-\mathrm{cosec}x+C$∫ 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: $\tan x=\frac{\sin x}{\cos x}$tan x = (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: $\int \tan xdx=-\ln |\cos x|+C$∫ tan x dx = −ln|cos x| + C. Check: −ln|cos x| differentiates to $-\frac{-\sin x}{\cos x}=\tan x$−(−sin x)/(cos x) = tan x.

The integral of tan x from 0 to $\frac{\pi }{3}$π/3 is $-\ln \left(\cos \left(\frac{\pi }{3}\right)\right)+\ln \left(\cos 0\right)=-\ln \frac{1}{2}+0=\ln 2\approx 0.693$−ln(cos(π/3)) + ln(cos 0) = −ln ½ + 0 = ln 2 ≈ 0.693.

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

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

Two answers that are one

Some books give $\int \tan xdx=\ln |\sec x|+C$∫ 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 $-\ln a=\ln \left(1/a\right)$−ln a = ln(1 / a). So $-\ln |\cos x|=\ln |1/\cos x|$−ln|cos x| = ln|1 / cos x|, and $1/\cos x=\sec x$1 / cos x = sec x.

${\tan }^{2}x$tan²x and sec x

${\tan }^{2}x$tan²x is not a derivative either, but the identity $1+{\tan }^{2}x={\sec }^{2}x$1 + tan²x = sec²x turns it into one: ${\tan }^{2}x={\sec }^{2}x-1$tan²x = sec²x − 1, so $\int {\tan }^{2}xdx=\tan x-x+C$∫ tan²x dx = tan x − x + C. From 0 to $\frac{\pi }{4}$π/4 that is $1-\frac{\pi }{4}\approx 0.215$1 − π/4 ≈ 0.215.

sec x itself needs a trick. Multiply it by $\frac{\sec x+\tan x}{\sec x+\tan x}$(sec x + tan x)/(sec x + tan x), which is 1: the integrand becomes $\frac{{\sec }^{2}x+\sec x\tan x}{\sec x+\tan x}$(sec²x + sec x tan x)/(sec x + tan x). The numerator is exactly the derivative of the denominator, so $\int \sec xdx=\ln |\sec x+\tan x|+C$∫ sec x dx = ln|sec x + tan x| + C. From 0 to $\frac{\pi }{4}$π/4 that is $\ln \left(\sqrt{2}+1\right)-\ln 1\approx 0.881$ln(√2 + 1) − ln 1 ≈ 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 ${\sec }^{2}x$sec²x.

Mixing up the family. ${\sec }^{2}x$sec²x 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 $\int \tan xdx=-\ln |\cos x|+C$∫ 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 $\int \cot xdx=\ln |\sin x|+C$∫ 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.