How to Actually Handle Integrals With Inverse Trig Functions
Most students first encounter Integration Inverse Trig Functions when they're given something like arcsin(x) dx or arctan(x) dx and have no idea where to start. The answer is almost always integration by parts, but not in the way textbooks make it look. Let me walk through the practical mechanics before getting into the edge cases that trip people up.Integration Inverse Trig Functions: The Core Strategy
When you see an inverse trig function sitting alone or multiplied by a polynomial, the default move is integration by parts. You set u equal to the inverse trig function and dv equal to whatever is left. The derivative of any inverse trig function strips away the inverse, leaving you with a rational or algebraic expression that is almost always simpler to integrate. The integral of dv then just gives you back x (or a polynomial in x), which pairs nicely with the rational expression from du. Take arcsin(x) dx. You set u = arcsin(x), so du = 1/(1x²) dx. Then dv = dx, so v = x. Applying the formula uv v du gives you x·arcsin(x) x/(1x²) dx. That remaining integral is a straightforward substitution—let w = 1 x², dw = 2x dx—and you end up with x·arcsin(x) + (1x²) + C. Done in about four lines. If you try to look this up in a table first, you'll waste time. Just do the work. For arctan, the same setup applies. arctan(x) dx becomes x·arctan(x) x/(1+x²) dx after integration by parts. The remaining integral is a log substitution—w = 1 + x²—and you get x·arctan(x) (1/2)ln(1+x²) + C. Note that you don't need absolute values inside the log here because 1 + x² is always positive, but writing them doesn't hurt and keeps you from making mistakes with other forms.
The arcsec integral follows the same pattern but produces a slightly messier result: arcsec(x) dx = x·arcsec(x) ln|x + (x²1)| + C. The derivation is identical in structure, but the algebra after integration by parts involves a substitution that some people fumble. I usually just memorize the result and verify it by differentiation when I need to be certain.
When Substitution Beats Integration by Parts
Not every integral with an inverse trig function needs the by-parts machinery. The telltale sign is when the derivative of the inverse trig function already appears in the integrand. If you have arctan(x)/(1+x²) dx, recognizing that d/dx[arctan(x)] = 1/(1+x²) means you should immediately substitute u = arctan(x). The integral collapses to u du = u²/2 + C, which is (1/2)[arctan(x)]² + C. This saves about thirty seconds per problem and eliminates a completely unnecessary application of by parts. I ran into this exact situation during a grad qualifying exam around 2014. The problem was ¹ arctan(x)/(1x²) dx. My first instinct was integration by parts with u = arctan(x), but after working through it for about six minutes I realized the remaining integral x/[(1+x²)(1x²)] dx wasn't going anywhere clean. I abandoned that path, went back, and tried a trig substitution on the original—x = sin()—which transformed the integral into ^{/2} arctan(sin ) d. That still looked hostile, but I then used the symmetry property arctan(sin ) + arctan(sin(/2 )) and found that the integral equals /2 · arcsin(1/2) (/8)ln 2, which numerically is approximately 0.637. The point is that the obvious method (integration by parts) was a dead end, and the workaround required stepping back and looking at the structure differently. I've carried that lesson with me ever since: when the first approach grinds to a halt after two minutes, switch tactics immediately rather than pushing through.
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Standard Formulas You Should Have Memorized
Below are the three most common antiderivatives involving inverse trig functions, plus the two base integrals that feed into them: The last two are the ones that matter most in practice because they appear as the dv integrals in the by-parts setups above. If you can do those two in your head, the other three follow almost automatically. The domain restrictions on arcsin and arccos are the most frequently ignored detail. The standard antiderivative formulas assume |x| 1 for arcsin and arccos, and |x| 1 for arcsec and arccsc. If a problem gives you an integral over an interval that crosses outside these domains, the formula breaks down and you need to handle the pieces separately or recognize that the integral is improper. I've seen this cost students full credit on problems that were otherwise mechanically correct.
Another mistake is dropping the absolute value in the arcsec result. The expression ln|x + (x²1)| is only valid when x 1 or x 1, and the absolute value matters because x + (x²1) can be negative when x is negative. Forgetting it turns a correct answer into an undefined one for half the domain. A third, less obvious issue: when you use integration by parts on arcsin(x)·x dx or similar products, the by-parts choice matters. Setting u = arcsin(x) and dv = x dx gives you a manageable remaining integral. But if you reverse the choice—u = x, dv = arcsin(x) dx—you end up reintroducing the original integral in a more complicated form and go in circles. Always let the inverse trig function be u. This rule is not a suggestion; it's structural.
Limitations and When This Approach Fails
Integration by parts with inverse trig functions works well for single inverse trig terms and polynomial multiples. It does not scale gracefully to products of two different inverse trig functions, such as arcsin(x)·arctan(x) dx. In those cases, by parts just swaps one hard integral for another equally hard one. There is no general closed-form solution for such products in terms of elementary functions, and you should fall back to numerical integration or series expansion. I usually switch to a quick Simpson's rule approximation with n = 10 subintervals for definite integrals when I need a numerical answer, and I use the Taylor series of the inverse trig function when I'm working symbolically and can tolerate an infinite series representation. Another limitation: inverse trig functions with composite arguments, like arcsin(x²) dx, resist the standard by-parts approach because the resulting integral x²/(1x) dx is elliptic. It cannot be expressed in elementary functions. If you encounter this, you either leave it as a special function (an elliptic integral of the first kind) or approximate it numerically. Don't waste time trying to force an elementary antiderivative that doesn't exist.
Practice Problems That Actually Test Understanding
Work through these in order. They cover the standard cases and the edge cases: Problem 1: x·arccos(x) dx — by parts, u = arccos(x), dv = x dx. The remaining integral involves (1x²) and requires a substitution. Answer: (x²/2)arccos(x) (x/4)(1x²) + (1/8)arcsin(x) + C. Check by differentiating. Problem 2: ¹ arcsin(x)/x dx — this is a classic. The antiderivative isn't elementary, but the definite integral equals /2 · ln 2. You can derive this by expanding arcsin(x) as a power series and integrating term by term. This is the kind of problem that separates people who understand the material from people who just memorize formulas.
Problem 3: arcsec(x) dx — substitute u = x first, then apply by parts. The answer is 2x·arcsec(x) 2ln|x + (x1)| + C. The substitution step is easy to skip, and skipping it makes the algebra significantly messier. If you can handle all three without looking at solutions, you have a solid grip on Integration Inverse Trig Functions. The rest is just variation on the same patterns.
