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integrals of inverse trig functions — Definition, Formula & Examples

Integrals of inverse trig functions are antiderivative formulas for expressions like arcsin(x), arccos(x), and arctan(x). These integrals are typically solved using integration by parts, since inverse trig functions have known derivatives but no direct antiderivative rules.

The indefinite integrals of the six inverse trigonometric functions are obtained by applying integration by parts with the inverse trig function as the factor to differentiate. For example, ∫arctan⁡(x) dx=xarctan⁡(x)−12ln⁡(1+x2)+C\int \arctan(x)\,dx = x\arctan(x) - \frac{1}{2}\ln(1+x^2) + C. Each result combines an algebraic-times-inverse-trig term with a logarithmic or radical correction term.

Key Formula

∫arcsin⁡(x) dx=xarcsin⁡(x)+1−x2+C\int \arcsin(x)\,dx = x\arcsin(x) + \sqrt{1-x^2} + C ∫arccos⁡(x) dx=xarccos⁡(x)−1−x2+C\int \arccos(x)\,dx = x\arccos(x) - \sqrt{1-x^2} + C ∫arctan⁡(x) dx=xarctan⁡(x)−12ln⁡(1+x2)+C\int \arctan(x)\,dx = x\arctan(x) - \tfrac{1}{2}\ln(1+x^2) + C
Where:
  • xx = The variable of integration
  • CC = Constant of integration

How It Works

To integrate an inverse trig function, choose it as uu in integration by parts (since its derivative is algebraic and simpler). The remaining dvdv is typically dxdx. After applying the formula ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du, you are left with a rational or radical integral that can be evaluated with standard techniques like substitution.

Worked Example

Problem: Evaluate the integral: ∫ arcsin(x) dx.
Set up by parts: Let u = arcsin(x) and dv = dx. Then du = 1/√(1−x²) dx and v = x.
u=arcsin⁡(x),dv=dxu = \arcsin(x),\quad dv = dxdu=11−x2 dx,v=xdu = \frac{1}{\sqrt{1-x^2}}\,dx,\quad v = x
Apply integration by parts: Use the formula ∫ u dv = uv − ∫ v du.
∫arcsin⁡(x) dx=xarcsin⁡(x)−∫x1−x2 dx\int \arcsin(x)\,dx = x\arcsin(x) - \int \frac{x}{\sqrt{1-x^2}}\,dx
Evaluate the remaining integral: Substitute w = 1 − x², so dw = −2x dx. The integral becomes (1/2)∫ w^(−1/2) dw = √w = √(1−x²).
−∫x1−x2 dx=1−x2-\int \frac{x}{\sqrt{1-x^2}}\,dx = \sqrt{1-x^2}
Answer: ∫arcsin⁡(x) dx=xarcsin⁡(x)+1−x2+C\int \arcsin(x)\,dx = x\arcsin(x) + \sqrt{1-x^2} + C

Why It Matters

These integrals appear regularly in AP Calculus BC and college-level calculus courses. They also arise in physics and engineering when modeling angles, arc lengths, or signal processing problems that involve inverse trig expressions.

Common Mistakes

Mistake: Confusing integrals OF inverse trig functions with integrals that PRODUCE inverse trig functions (e.g., ∫ 1/√(1−x²) dx = arcsin(x) + C).
Correction: Integrals of inverse trig functions have arcsin(x), arctan(x), etc. in the integrand. Integrals producing inverse trig results have algebraic/radical integrands. Know which direction you are working in.

Related Terms