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Inverse Function Integration — Definition, Formula & Examples

Inverse function integration is a technique that expresses the integral of an inverse function f−1(x)f^{-1}(x) in terms of the integral of the original function ff. It relies on the geometric fact that the graph of f−1f^{-1} is the reflection of the graph of ff across the line y=xy = x.

If ff is a continuous, strictly monotonic function on [a,b][a, b] with inverse f−1f^{-1}, then ∫f−1(y) dy=y f−1(y)−F(f−1(y))+C\int f^{-1}(y)\,dy = y\,f^{-1}(y) - F(f^{-1}(y)) + C, where FF is an antiderivative of ff. Equivalently, for definite integrals: ∫f(a)f(b)f−1(y) dy=b f(b)−a f(a)−∫abf(x) dx\int_{f(a)}^{f(b)} f^{-1}(y)\,dy = b\,f(b) - a\,f(a) - \int_a^b f(x)\,dx.

Key Formula

∫f−1(y) dy=y f−1(y)−F ⁣(f−1(y))+C\int f^{-1}(y)\,dy = y\,f^{-1}(y) - F\!\left(f^{-1}(y)\right) + C
Where:
  • f−1f^{-1} = The inverse of the function f
  • FF = An antiderivative of f, i.e., F'(x) = f(x)
  • yy = The variable of integration
  • CC = Constant of integration

How It Works

The formula comes from integration by parts. Set u=f−1(y)u = f^{-1}(y) and dv=dydv = dy, so du=1f′(f−1(y)) dydu = \frac{1}{f'(f^{-1}(y))}\,dy and v=yv = y. After substituting x=f−1(y)x = f^{-1}(y), the remaining integral converts back to an integral of f(x)f(x), which is often much easier to evaluate. Geometrically, the area under f−1f^{-1} from f(a)f(a) to f(b)f(b) and the area under ff from aa to bb together tile the rectangle with corners (0,0)(0,0), (b,0)(b,0), (b,f(b))(b,f(b)), and (0,f(b))(0,f(b)), minus the rectangle involving aa and f(a)f(a).

Worked Example

Problem: Evaluate ∫01arcsin⁡(y) dy\int_0^1 \arcsin(y)\,dy.
Identify f and its inverse: Here f(x)=sin⁡(x)f(x) = \sin(x), so f−1(y)=arcsin⁡(y)f^{-1}(y) = \arcsin(y). An antiderivative of sin⁡(x)\sin(x) is F(x)=−cos⁡(x)F(x) = -\cos(x). The limits y=0y = 0 to y=1y = 1 correspond to x=0x = 0 to x=π/2x = \pi/2.
Apply the definite integral formula: Use ∫f(a)f(b)f−1(y) dy=b f(b)−a f(a)−∫abf(x) dx\int_{f(a)}^{f(b)} f^{-1}(y)\,dy = b\,f(b) - a\,f(a) - \int_a^b f(x)\,dx with a=0a = 0 and b=π/2b = \pi/2.
∫01arcsin⁡(y) dy=π2⋅sin⁡ ⁣(π2)−0⋅sin⁡(0)−∫0π/2sin⁡(x) dx\int_0^1 \arcsin(y)\,dy = \frac{\pi}{2}\cdot\sin\!\left(\frac{\pi}{2}\right) - 0\cdot\sin(0) - \int_0^{\pi/2} \sin(x)\,dx
Evaluate: Compute each piece: π2⋅1=π2\frac{\pi}{2} \cdot 1 = \frac{\pi}{2} and ∫0π/2sin⁡(x) dx=[−cos⁡(x)]0π/2=0−(−1)=1\int_0^{\pi/2} \sin(x)\,dx = [-\cos(x)]_0^{\pi/2} = 0 - (-1) = 1.
∫01arcsin⁡(y) dy=π2−1\int_0^1 \arcsin(y)\,dy = \frac{\pi}{2} - 1
Answer: π2−1\dfrac{\pi}{2} - 1

Why It Matters

This technique is especially useful for integrating inverse trigonometric functions like arcsin⁡\arcsin, arctan⁡\arctan, and arccos⁡\arccos, as well as logarithms (since ln⁡\ln is the inverse of exp⁡\exp). It appears frequently in Calculus II courses and provides an elegant alternative when direct antidifferentiation of the inverse function is difficult.

Common Mistakes

Mistake: Forgetting to adjust the limits of integration when switching between f and its inverse.
Correction: If integrating f−1f^{-1} from cc to dd, the corresponding limits for ff are f−1(c)f^{-1}(c) to f−1(d)f^{-1}(d). Always verify that the limits match using y=f(x)y = f(x).