Integration Of Inverse Trigonometric Functions

Integration By Parts For Inverse Trig Functions

Most students encounter Integration Of Inverse Trigonometric Functions as a standalone chapter in a calculus course. The technique itself is routine — it's always integration by parts — but the way it's taught glosses over the things that actually go wrong when you're working problems under time pressure. I'll walk through the mechanics, the edge cases, and what I learned after spending years watching students (and junior colleagues) burn minutes on simple mistakes.

Integration Of Inverse Trigonometric Functions: The Standard Formulas

The core idea is that every inverse trig integral reduces to the same pattern. You set u equal to the inverse trig function and dv equal to dx. That gives you v = x, and the new integral you're left with is always algebraic. Here are the six standard results: arcsin(x)dx = x·arcsin(x) + (1-x²) + C arccos(x)dx = x·arccos(x) - (1-x²) + C

arctan(x)dx = x·arctan(x) - ½ln(1+x²) + C arccot(x)dx = x·arccot(x) + ½ln(1+x²) + C arcsec(x)dx = x·arcsec(x) - ln|x + (x²-1)| + C

arccsc(x)dx = x·arccsc(x) + ln|x + (x²-1)| + C You don't need to memorize all six from scratch. Deriving any one of them takes about forty seconds, and since they all use the same method, deriving one teaches you the rest.

The Method In Practice

Let me walk through the arcsin case, which is the one most people get wrong on the sign. Let I = arcsin(x)dx. Set u = arcsin(x), so du = 1/(1-x²) dx. Set dv = dx, so v = x. By parts: I = x·arcsin(x) - x/(1-x²) dx

That remaining integral looks ugly but it's a simple substitution. Let w = 1-x², dw = -2x dx, so x dx = -dw/2. The integral becomes -(-1/2)·w^(-½) dw, which evaluates to +(1-x²). Putting it together: I = x·arcsin(x) + (1-x²) + C The sign trip-up is real. The integral of x/(1-x²) gives you -(1-x²), and since you're subtracting that integral, the minus cancels to plus. I've lost points on this exact sign error more than once during grading.

For arctan, the algebra is slightly different but follows the same template. u = arctan(x), du = 1/(1+x²) dx, v = x. The remaining integral is x/(1+x²) dx, which is a direct substitution w = 1+x², giving ½ln(1+x²). The final result has a minus sign because you subtract the integral.

The Substitution Problem That Actually Trips People Up

Here's a case I dealt with recently that doesn't appear in standard textbooks. You're asked to integrate arcsin(2x/(1+x²)). At first glance this looks like it needs a messy integration by parts, but it simplifies dramatically if you recognize the composition. Let x = tan(). Then 2x/(1+x²) = 2tan()/sec() = 2sin()cos() = sin(2). So arcsin(2x/(1+x²)) = arcsin(sin(2)) = 2 = 2arctan(x), valid on the appropriate domain. This reduces the entire integral to 2arctan(x)dx, which is just twice the standard formula. I encountered this exact problem in a graduate-level quals exam prep sheet. The intended solution path wasn't brute-force by parts — it was recognizing the trigonometric identity hidden inside the inverse function argument. Textbooks don't always flag that these compositions can collapse nicely, and students who don't see it will spend eight minutes on an integral that takes forty seconds with the right substitution.

A Harder Edge Case: When By Parts Creates a Loop

Sometimes integration by parts produces an equation you need to solve for I. Consider arctan(x)dx. Substitute u = x so x = u² and dx = 2u du. You now have 2u·arctan(u)du. Apply parts: let U = arctan(u), dV = 2u du. Then dU = 1/(1+u²) du and V = u². The new integral is u²/(1+u²) du, which looks like it should be polynomial long division. It is: u²/(1+u²) = 1 - 1/(1+u²). That integrates to u - arctan(u). Going back through the substitution chain gets you the answer. The trick here isn't circular — it's recognizing that the by-parts result decomposes into two pieces, one of which cancels part of what you already have. This type of problem shows up frequently in second-semester calculus. The pattern to watch for: after your first by-parts step, if the remaining integral is rational and nearly cancels with something you already produced, you're on the right track. If it looks like it's going in circles without simplification, you either picked the wrong u/dv split or you're missing an algebraic simplification in the new integral.

Common Pitfalls With Domain And Absolute Values

The arcsec and arccsc integrals carry absolute value signs for a reason. The expression (x²-1) is only real for |x| 1, and the logarithm term ln|x + (x²-1)| handles both positive and negative domains. If you drop the absolute value, your antiderivative is only valid for x > 1 and will give you complex results for x

-1. Another practical issue: when you're doing definite integrals involving arcsec or arccsc, you need to check that the interval of integration actually lies within the domain. arcsec(x) is undefined on (-1, 1). I once saw a student integrate arcsec(x) from -2 to 2 without noting that the function doesn't exist between -1 and 1. The integral is improper at those points and diverges. Worth checking before you start.

When The Formula Doesn't Help And What To Do Instead

Not every integral involving inverse trig functions can be handled with the standard by-parts formula. If your integrand is something like x²·arcsin(x), the first by-parts step reduces the power of x but leaves you with a harder integral. You apply by parts again — this time with u = arcsin(x) and dv = x²dx — and you end up with x²/(1-x²) dx. That one requires a trig substitution x = sin(), which turns it into sin²() d. That's manageable but it's three steps instead of one. If you're dealing with products like arcsin(x)·arccos(x) or arctan(x)·arcsec(x), there's no general shortcut. You'd need to apply identities first — for instance, arcsin(x) + arccos(x) = /2 — to reduce the expression before attempting integration. Skipping this simplification step is probably the most common waste of time I see. Students jump straight into by parts on a product that could be reduced to a constant plus a single inverse trig function in one line.

Numerical Integration As A Fallback

For integrals that resist closed-form solutions — say, arcsin(x²)dx over a specific interval — numerical methods are the practical answer. Gaussian quadrature gives excellent accuracy with relatively few evaluation points for smooth integrands like these. If you're working in engineering and need a numeric result quickly, adaptive Simpson's rule is straightforward to implement and converges fast for inverse trig functions because they're smooth on their domains (away from the branch points at ±1 for arcsin and arccos). The tradeoff is that you lose the exact form, which matters if you're building a larger symbolic expression. But for most applied work, a numeric result to six decimal places is sufficient and takes seconds rather than minutes of algebra.

Integration Of Inverse Trigonometric Functions: What To Remember

The method is always integration by parts with u = the inverse trig function. The resulting integral is always algebraic and usually solvable with a single substitution or a basic trig substitution. The things that make these problems hard aren't the formulas — they're the compositions that hide simplifications, the domain restrictions you forget to check, and the cases where the by-parts loop requires algebraic resolution instead of just writing down an answer. If you want a reference sheet with all six formulas and worked examples, I keep a personal pdf at https://example.com/integration-inverse-trig-reference. It includes the loop-resolving technique and the arcsec domain discussion I mentioned above. No subscription required.