Integrating The Inverse Sine And Cosine

Why These Integrals Are Useful

Inverse trigonometric functions have derivatives involving a square root. That derivative may look complicated, but in an integral such as \(\int \arcsin x\,dx\), integration by parts turns the inverse function into a simpler algebraic expression. The same strategy works for \(\arccos x\), provided its negative derivative is handled carefully.

The key is to choose the inverse trigonometric function as the factor to differentiate and \(1\) as the factor to integrate. This produces an integral involving \(x/\sqrt{1-x^2}\), which has a simple antiderivative.

Inverse Sine and Inverse Cosine

Here, \(\arcsin x\) denotes the principal inverse sine, with values in \([-\pi/2,\pi/2]\), and \(\arccos x\) denotes the principal inverse cosine, with values in \([0,\pi]\). Both functions have domain \([-1,1]\). On the open interval \((-1,1)\), their derivatives are

The square root is the nonnegative square root. The minus sign in the derivative of \(\arccos x\) is essential: as \(x\) increases, the principal angle \(\arccos x\) decreases.

Integration by parts states that \(\int u\,dv=uv-\int v\,du\). For these integrals, take \(u\) to be the inverse trigonometric function and \(dv=dx\), so \(v=x\). This choice moves the inverse function into its derivative and leaves an integral that can be evaluated directly.

Deriving the Inverse Sine Antiderivative

For \(-1<x<1\), choose \(u=\arcsin x\) and \(dv=dx\). Then \(du=dx/\sqrt{1-x^2}\) and \(v=x\). Applying integration by parts gives

To evaluate the remaining integral, note that the derivative of \(\sqrt{1-x^2}\) is \(-x/\sqrt{1-x^2}\). Therefore, an antiderivative of \(x/\sqrt{1-x^2}\) is \(-\sqrt{1-x^2}\). Substituting this into the integration-by-parts result yields

The plus sign before the square root appears because integration by parts subtracts the remaining integral, whose antiderivative already has a minus sign.

Deriving the Inverse Cosine Antiderivative

Now choose \(u=\arccos x\) and \(dv=dx\). The derivative of \(\arccos x\) has a minus sign, so \(du=-dx/\sqrt{1-x^2}\). Integration by parts gives

The remaining integral is \(-\sqrt{1-x^2}\), as above. Consequently,

The square-root term now has a minus sign. The integration-by-parts formula contributes a minus before the integral, but the negative derivative of \(\arccos x\) changes that to a plus before the remaining integral. That integral itself has a negative antiderivative.

Why the Method Works

The integration-by-parts step is a rearrangement of the product rule. In general, \((uv)'=u'v+uv'\). Integrating both sides and rearranging gives \(\int u\,dv=uv-\int v\,du\), when \(du=u'\,dx\) and \(dv=v'\,dx\). In these examples, choosing \(v=x\) makes the remaining product \(v\,du\) equal to either \(x/\sqrt{1-x^2}\,dx\) or its negative.

The simplification can also be understood through the chain rule. Differentiating \(\sqrt{1-x^2}\) produces the factor \(x/\sqrt{1-x^2}\), up to a minus sign. Thus the inverse-function derivative and the factor \(x\) created by \(v=x\) match the derivative of a simple square-root expression. Integration by parts is effective because it creates exactly this match.

A Relationship Between the Two Formulas

On \([-1,1]\), the principal inverse sine and inverse cosine satisfy \(\arcsin x+\arccos x=\pi/2\). Adding the two antiderivative expressions provides a useful consistency check: the square-root terms cancel, and the inverse-function terms combine according to that identity.

A Sign Error to Avoid

The inverse cosine formula is especially vulnerable to a sign error. For example, a tempting but incorrect expression is \(x\arccos x+\sqrt{1-x^2}\). Differentiate it before accepting it:

This is not \(\arccos x\) in general. The square-root terms reinforce one another rather than cancel. The corrected expression is \(x\arccos x-\sqrt{1-x^2}\), whose derivative is

A reliable check is to differentiate the entire proposed answer, including the square-root term. Checking only the inverse-function product can miss exactly this kind of error.

Domain and Constants

The displayed derivative formulas contain \(\sqrt{1-x^2}\) in the denominator, which is nonzero for \(-1<x<1\) and zero at \(x=\pm1\). The derivations therefore establish antiderivatives on the open interval. The original inverse functions are defined on the closed interval, and the expressions for the antiderivatives have finite endpoint values, but the derivative-based calculations above are made in the open interval.

As with every indefinite integral, the constant \(C\) represents an arbitrary real constant. Two antiderivatives of the same function on an interval differ by a constant. When combining several integrals, their constants can be absorbed into a single arbitrary constant.

The method matters because it links two kinds of information: the derivative of an inverse trigonometric function and the chain-rule structure of a square root. Remember the derivative signs, choose \(u\) to be the inverse function, and verify the result by differentiation.