Definite Integrals Involving The Inverse Sine

The Antiderivative of the Inverse Sine

Definite integrals of \(\arcsin x\) rest on a single antiderivative, which comes from integration by parts. Let \(u=\arcsin x\) and \(dv=dx\). Then \(du=\frac{dx}{\sqrt{1-x^2}}\) and \(v=x\), so \(\int\arcsin x\,dx=x\arcsin x-\int\frac{x}{\sqrt{1-x^2}}\,dx\). The remaining integral is solved by the substitution \(w=1-x^2\), \(dw=-2x\,dx\), which gives \(\int\frac{x\,dx}{\sqrt{1-x^2}}=-\sqrt{1-x^2}\).

This tutorial concentrates on evaluating the definite integral at endpoints strictly inside \((-1,1)\), where both \(\arcsin\) and \(\sqrt{1-x^2}\) take convenient exact values, and on explaining what the two endpoint terms mean.

Checking by Differentiation

The product rule gives \(\frac{d}{dx}[x\arcsin x]=\arcsin x+\frac{x}{\sqrt{1-x^2}}\). The derivative of \(\sqrt{1-x^2}\) is \(\frac{-x}{\sqrt{1-x^2}}\). The algebraic terms cancel, leaving \(\arcsin x\).

Exact Values to Keep Handy

Worked Examples

A Geometric Interpretation

The graph of \(y=\arcsin x\) on \([0,1]\) is the graph of \(x=\sin y\) on \([0,\frac\pi2]\), reflected across \(y=x\). The area under \(\arcsin x\) from \(0\) to \(1\) therefore equals the area of the rectangle \([0,1]\times[0,\frac\pi2]\) minus the area under \(\sin y\) from \(0\) to \(\frac\pi2\). That is \(\frac\pi2\cdot1-\int_0^{\pi/2}\sin y\,dy=\frac\pi2-1\), the same value found above. The term \(x\arcsin x\) in the antiderivative is the rectangle contribution at the upper endpoint, and \(\sqrt{1-x^2}\) accounts for the part cut off by the sine curve.

What the Endpoint Terms Mean

For a general interval \([a,b]\), the quantity \(b\arcsin b-a\arcsin a\) is the difference of two rectangle areas measured from the origin, while \(\sqrt{1-b^2}-\sqrt{1-a^2}\) is the change in the height of the unit semicircle. Since the semicircle falls as \(|x|\) grows, this second difference is negative for \(0\le a<b\le1\). The final area is a positive number because the rectangle term outweighs the semicircle drop.

A Reliable Procedure

Common Mistakes

The most common error is to drop the term \(\sqrt{1-x^2}\) or its sign, writing \(x\arcsin x-\sqrt{1-x^2}\). Differentiating shows this is wrong: its derivative is \(\arcsin x+\frac{2x}{\sqrt{1-x^2}}\). A second error is to evaluate a symmetric interval \([-b,b]\) as \(2F(b)\) instead of \(F(b)-F(-b)\). Because \(F\) is even, the correct difference is zero, and the shortcut \(2F(b)\) would produce a large false value.

Practice Problems with Answers

Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.

1. Evaluate \(\int_0^{1/2}\arcsin x\,dx\). Hint: use the table of exact values. Answer: \(\frac\pi{12}+\frac{\sqrt3}{2}-1\)
2. Evaluate \(\int_0^1\arcsin x\,dx\). Hint: the radical vanishes at \(x=1\). Answer: \(\frac\pi2-1\)
3. Evaluate \(\int_{1/2}^{\sqrt3/2}\arcsin x\,dx\). Hint: subtract \(F(\frac12)\) from \(F(\frac{\sqrt3}{2})\). Answer: \(\frac{\pi\sqrt3}{6}-\frac\pi{12}+\frac12-\frac{\sqrt3}{2}\)
4. Find an antiderivative of \(2\arcsin x\). Hint: double the basic result. Answer: \(2x\arcsin x+2\sqrt{1-x^2}\)

Interpreting the Result

Because \(\arcsin x\) is odd, the signed area on \([-b,0]\) is the negative of the area on \([0,b]\). The antiderivative \(F(x)=x\arcsin x+\sqrt{1-x^2}\) is even, with its minimum value \(F(0)=1\) at the origin. This is consistent: \(F'(x)=\arcsin x\) is negative for \(x<0\) and positive for \(x>0\), so \(F\) decreases to the origin and increases away from it. Any definite integral is simply the change in \(F\), so integrals that start and end at points with the same \(|x|\) give zero.

A useful physical picture is a particle whose velocity is \(\arcsin t\). Its displacement over \([0,1]\) is \(\frac\pi2-1\approx0.571\), while over \([-1,1]\) it returns to its starting position because the negative and positive velocities cancel. The total distance travelled over \([-1,1]\), which uses the absolute value of the velocity, is twice the displacement, namely \(\pi-2\).

Improper-Looking Endpoints Are Not Improper

The derivative \(\frac{1}{\sqrt{1-x^2}}\) is unbounded at \(x=\pm1\), so it is tempting to worry that integrals of \(\arcsin x\) up to \(1\) are improper. They are not. The integrand \(\arcsin x\) itself is continuous on \([-1,1]\), with value \(\frac\pi2\) at \(x=1\). Only the integrand of the intermediate step, \(\frac{x}{\sqrt{1-x^2}}\), is unbounded, and that integral still converges because its antiderivative \(-\sqrt{1-x^2}\) has a finite limit. The final formula therefore applies directly at \(x=\pm1\) with no limits required.

A Numerical Spot Check

A quick estimate protects against slips with exact values. For \(\int_0^{1/2}\arcsin x\,dx\), apply the trapezoid rule with a single interval: the heights are \(0\) and \(\frac\pi6\approx0.5236\), so the estimate is \(\frac12\cdot\frac{0+0.5236}{2}=0.1309\). The exact value is \(0.1278\). The estimate is slightly high because \(\arcsin x\) is convex on \([0,1]\), so the chord lies above the curve. Agreement to two digits is what one should see from such a coarse rule, and it confirms both the antiderivative and the arithmetic with \(\frac\pi{12}\) and \(\frac{\sqrt3}{2}\).

Make this habit routine: whenever an exact answer involves several special values, compare it to a chord or rectangle estimate. A mismatch of more than a few percent usually means a wrong sign or a misread special angle.

Where This Fits in the Course

This tutorial belongs to the module “Integration by Parts I”. The tutorial just before it is “Definite Integrals Involving The Inverse Tangent,” and the one that follows is “Domain And Endpoint Checks For Inverse-Trigonometric Parts.” The reflection idea used for the geometric check returns when integrals of inverse functions are studied directly, and the exact-value arithmetic here is practice for the trigonometric substitutions that follow.