Integrals Of Inverse Trig Functions

Getting the Basics Straight

The standard Integrals Of Inverse Trig Functions rely on integration by parts. You pick the inverse trig function as your u and the rest as dv. That part is straightforward. The harder part is recognizing when the formula actually applies and when you need to manipulate the integrand first. Here are the three you'll use most: arctan(x) dx = x·arctan(x) - (1/2)·ln(1 + x²) + C

arcsin(x) dx = x·arcsin(x) + (1 - x²) + C arccos(x) dx = x·arccos(x) - (1 - x²) + C The arccot, arcsec, and arccsc versions exist but show up less frequently in practice. I'll cover them if needed, but the three above handle the vast majority of real problems.

Integrals Of Inverse Trig Functions: The Method

Integration by parts gives you the framework. Take arctan(x)·f(x) dx as your template. Set u = arctan(x), so du = 1/(1 + x²) dx. Set dv = f(x) dx, so v is whatever antiderivative you can find. Then you're left with v/(1 + x²) dx, which is often simpler than the original. But here's the thing that trips people up: the formula arctan(x) dx = x·arctan(x) - (1/2)·ln(1 + x²) + C is itself the result of applying integration by parts once. You prove it by letting u = arctan(x) and dv = 1 dx. Then you hit x/(1 + x²) dx, which is a basic u-substitution. Don't skip that step mentally. When you're doing this under time pressure, your brain will want to jump straight to the memorized result, and that's where sign errors creep in. For arcsin(x) dx, the by-parts setup gives du = 1/(1 - x²) dx and v = x. The resulting integral is x/(1 - x²) dx, which solves cleanly with the substitution w = 1 - x². Same pattern. The arccos version follows identically except the sign flips because d/dx[arccos(x)] = -1/(1 - x²).

When the Integrand Isn't Just the Inverse Trig Function

This is where things get interesting. Most problems you'll encounter don't look like the clean textbook examples. They look like x²·arctan(x) dx or (1 - x²)·arcsin(x) dx or something equally uninviting. Take x²·arctan(x) dx. Apply by-parts with u = arctan(x), dv = x² dx. You get (1/3)x³·arctan(x) - (1/3)x³/(1 + x²) dx. Now you need to handle x³/(1 + x²) dx. Polynomial long division or algebraic rearrangement: x³/(1 + x²) = x - x/(1 + x²). That splits into x dx - x/(1 + x²) dx. The first is (1/2)x². The second is (1/2)·ln(1 + x²). Put it all together and you get: (1/3)x³·arctan(x) - (1/6)x² + (1/6)·ln(1 + x²) + C

I worked through a similar problem recently involving x³·arctan(x²) dx over [0, 1]. The substitution u = x² made it (1/2)u·arctan(u) du, which then went straight to the standard formula. Without that first substitution, I would have spent twice as long and made at least one algebra mistake. The lesson: always check for a substitution that simplifies the argument of the inverse trig function before reaching for by-parts.

A Real Problem That Broke the Standard Approach

Last year I was evaluating ¹ ln(x)·arcsin(x) dx. This doesn't fit any standard formula. The logarithm makes direct by-parts messy, and substitution doesn't help either. I ended up expanding arcsin(x) as its Taylor series, integrating term by term, and then summing the resulting series numerically. It took about 40 minutes of work and gave me approximately -0.3298. The closed form involves polylogarithms and isn't particularly illuminating. The takeaway: not every integral involving inverse trig functions has a clean elementary antiderivative. If you've applied by-parts twice and you're still stuck, the integral might simply not yield to standard techniques. You need a different strategy or you need to accept a numerical answer.

Common Pitfalls That Waste Hours

The first one is domain. arcsin(x) and arccos(x) are only defined on [-1, 1]. If your integral has bounds outside that range, the integrand is undefined and you're dealing with complex values. This comes up more often than you'd think in applied problems where the bounds come from a physical constraint. The second is the derivative sign. d/dx[arcsin(x)] = 1/(1 - x²). d/dx[arccos(x)] = -1/(1 - x²). Get the sign wrong in your by-parts setup and the entire result flips. I've seen this error in exam solutions and it's embarrassing how often it happens. The third is forgetting the absolute value in logarithmic terms. The integral of 1/(1 + x²) after by-parts sometimes produces ln|expression|. If you drop the absolute value, you'll get incorrect results for negative arguments. This matters especially in definite integrals where the argument crosses zero.

Advanced Nuance: Substitution Before By-Parts

Here's something most textbooks don't emphasize enough. Sometimes the fastest path isn't by-parts first. It's substitution first, then by-parts. Consider arctan(x) dx. Direct by-parts gives you u = arctan(x) and dv = dx. Then du = 1/(2x(1 + x)) dx. That denominator is ugly. But if you substitute u = x first, you get 2arctan(u)·u du, which is a much cleaner by-parts problem. The substitution transformed an awkward integral into a standard one in two steps. Similarly, arcsin(x/2) dx benefits from the substitution u = x/2, giving 2arcsin(u) du. The formula applies directly after that. Skipping the substitution saves you nothing and costs you clarity.

When This Method Completely Fails

Definite integrals over infinite domains are the main failure case. ^ arctan(x)/x² dx converges, but the antiderivative you get from by-parts involves terms that blow up at infinity. You need to evaluate the limit carefully, and even then you might find that the improper integral requires contour integration or another advanced technique for a clean result. The standard formula gives you the machinery, but it doesn't guarantee the machinery produces a usable answer for every bound configuration. Another failure mode: when the integrand contains inverse trig functions composed with each other, like arcsin(arctan(x)). No elementary antiderivative exists for that. You'd need special functions or numerical integration. Recognizing this early saves you from wasting time on a dead end.

Quick Reference: All Six Core Formulas

arcsin(x) dx = x·arcsin(x) + (1 - x²) + C arccos(x) dx = x·arccos(x) - (1 - x²) + C arctan(x) dx = x·arctan(x) - (1/2)·ln(1 + x²) + C

arccot(x) dx = x·arccot(x) + (1/2)·ln(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

Memorize the first three. The last three are derivable from them or from their co-function relationships. If you understand the by-parts pattern, you can reconstruct any of these in under a minute during an exam.

Worked Example: Definite Integral

Compute ¹ x²·arctan(x) dx. By-parts: u = arctan(x), dv = x² dx. du = dx/(1 + x²), v = x³/3. ¹ x²·arctan(x) dx = [x³/3·arctan(x)]¹ - ¹ x³/(3(1 + x²)) dx

First term: (1/3)·arctan(1) - 0 = /12. Second integral: ¹ x³/(1 + x²) dx = ¹ (x - x/(1 + x²)) dx = [x²/2 - (1/2)·ln(1 + x²)]¹ = 1/2 - (1/2)·ln(2). Combine: /12 - (1/3)(1/2 - (1/2)·ln(2)) = /12 - 1/6 + (1/6)·ln(2) 0.108.

That's the complete answer. No shortcuts, no mysteries. Just careful bookkeeping through the algebra.