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.
Factor x out of the radical and substitute t=1+x41; the integral collapses to a single power rule and evaluates to −(x4x4+1)1/4+c, which is option (D).
The integrand consistent with the answer choices is x2(x4+1)3/41 (the radical exponent that makes one of the options an exact antiderivative).
Factor x out of the radical.
(x4+1)3/4=[x4(1+x41)]3/4=x3(1+x41)3/4,
so
∫x2(x4+1)3/4dx=∫x5(1+x41)3/4dx.
Substitutet=1+x41. Then dt=−x54dx, i.e. x5dx=−41dt:
The integral is solved by substituting t=cosx, which turns it into a standard ∫a2+t2dt form. The correct answer is option (A).
The key insight here is that the numerator sinx is almost the derivative of cosx, up to a sign. Whenever you see sinx paired with a function of cosx, substitution t=cosx is the natural move — it collapses the integral into a rational function.
Let’s walk through it.
Substitutet=cosx. Then dt=−sinxdx, so sinxdx=−dt. The integral becomes
∫3+4cos2xsinxdx=∫3+4t2−dt.
Factor the denominator to match the standard form ∫u2+a2du=a1tan−1au+C. Write
Q.The curve passing through the point (1,2) given that the slope of the tangent at any point (x,y) is y2x represents
(A) Circle
(B) Parabola
(C) Ellipse
(D) Hyperbola
›Reveal solutionSolution
Separate the variables, integrate, fix the constant with the given point, and identify the resulting conic from the opposite signs of x2 and y2.
Step 1 — Translate the geometric condition into an ODE
"The slope of the tangent at any point (x,y) is y2x" means precisely
dxdy=y2x
Step 2 — Separate the variables and integrate
The equation is separable (all y's to one side, all x's to the other):