You already integrate sinx and cosx. Hyperbolic integration is the same idea with a different family: sinhx, coshx, tanhx, and their reciprocals.
The name comes from geometry: just as cost,sint trace a circle (x2+y2=1), cosht,sinht trace a hyperbola (x2−y2=1). The integration rules are almost identical to the trigonometric ones, with a few sign changes.
The core definitions
In terms of exponentials:
sinhx=2ex−e−x,coshx=2ex+e−x,tanhx=coshxsinhx
From these come the derivatives:
dxdsinhx=coshx,dxdcoshx=sinhx,dxdtanhx=sech2x
Notice the derivative of coshx is +sinhx (not −sinhx as in trigonometry). That plus sign is the only real difference from the circular case.
The integration formulas
Reversing the derivatives:
∫sinhxdx=coshx+C
∫coshxdx=sinhx+C
∫sech2xdx=tanhx+C
∫csch2xdx=−cothx+C
∫sechxtanhxdx=−sechx+C
∫cschxcothxdx=−cschx+C
Why the sign difference matters
Watch out
Don't treat ∫sinhxdx like ∫sinxdx. ∫sinxdx=−cosx+C, but ∫sinhxdx=+coshx+C — the minus sign is gone.
Check it: differentiate coshx and you get sinhx, not −sinhx, so the integral must be positive.
A worked example
Find ∫(3sinhx−2coshx)dx.
=3∫sinhxdx−2∫coshxdx=3coshx−2sinhx+C
When you use it in exams
Direct integration — apply the standard formulas above.
Substitution — a messy integral like ∫x2+a2dx becomes clean with x=asinht or x=acosht. That's a separate technique, but it relies on these basic integrals. …
The integral ∫ex+e−xdx simplifies by rewriting the denominator as 2coshx, then substituting t=ex to get a standard arctangent form. The correct answer is tan−1(ex)+C, which is option (A).
The key insight here is that the integrand ex+e−x1 looks like a hyperbolic secant function — because ex+e−x=2coshx, so the integrand is 21sech x. But the direct hyperbolic route isn't the simplest. Instead, notice that the denominator is symmetric in ex and e−x, which suggests a substitution that "breaks" this symmetry: let t=ex. This turns the integral into a rational function of t, which is a standard technique for integrals involving exponentials.
Rewrite the integrand
Multiply numerator and denominator by ex to clear the negative exponent:
∫ex+e−xdx=∫e2x+1exdx.
This step is crucial — it transforms the denominator into a simple quadratic in ex.
Substitute t=ex
Then dt=exdx, so the numerator exdx becomes exactly dt. The integral becomes:
Mistake 1: Guessing log(ex+e−x) by the ff′ pattern.
Why it's wrong: dxd(ex+e−x)=ex−e−x, not 1, so the numerator is not the denominator's derivative — the log form is wrong (that is option D, a distractor). Correct approach: multiply by ex and substitute t=ex.