How to Integrate Arcsin(x) or Sin-1(x)

Graphically, the inverse sine function looks like a small segment of the sine function reflected over the diagonal line y = x. In trigonometric terms, the function y = arcsin(x) or y = sin-1(x) means that if you input a number x between -1 and 1, the function will out put an angle measure y between 0 and π radians such that the sine of y equals x. In other words, the function y = arcsin(x) is equivalent to sin(y) = x.

To find the antiderivative of f(x) = arcsin(x) you must use a combination of integration by parts and u-substitution. Here the integral of arsin(x) is worked out step by step.

Graph of y = arcsin(x)

Integral of y = arcsin(x)

The first step to integrate the function f(x) = arcsin(x) is to apply the standard trick of integration by parts. The integration by parts formula is

∫ u * dv = u * v - ∫ v * du

In the integral ∫ arcsin(x) dx, we let u = arcsin(x), dx = dv, du = 1/sqrt(1 - x^2) dx, and v = x. This gives us the integral equation

∫ arcsin(x) dx = x*arcsin(x) - ∫ x/sqrt(1 - x^2) dx

The right-hand side of this equation contains a new integral that looks slightly complicated, but we can still work it out if we apply the trick of substitution. Let's set w = x^2 and dw = 2x dx. Now we get the integral equivalence

∫ x/sqrt(1 - x^2) dx
= (1/2) * ∫ 1/sqrt(1 - w) dw
= -sqrt(1 - w) + c
= -sqrt(1 - x^2) + c

If we put all the separate pieces together, we get the final expression for the antiderivative of the function y = arcsin(x).

∫ arcsin(x) dx
= x*arcsin(x) + sqrt(1 - x^2) + c

How to Use the Integral of arcsin(x)

The integral of the function y = arcsin(x) can be used to find the area under a curve, or to solve certain types of differential equations. The same steps above can also be used to integrate y = arccos(x), the inverse cosine, a closely related function. The integral of arccos(x) works out to be

∫ arccos(x) dx = x*arccos(x) - sqrt(1 - x^2) + c

Here are some more examples of how to use the integral of arcsin(x).

Example 1

What is the area of the region bounded by the y-axis, the line y = 0.5, and the curve y = sin(x)? The region is shaded in yellow in the graph below.

As a simple integral, the area of the region is best represented by

∫ arcsin(x) dx, on the interval 0 ≤ x ≤ 0.7

The area is then found by plugging the endpoints of the interval into the antederivative and subtracting. This gives us

[0.7*arcsin(0.7) + sqrt(1 - 0.49)] - [0*arcsin(0) + sqrt(1 - 0)]
= 0.256921

Example 2

Antiderivatives of functions are useful in find the general solutions to certain classes of differential equations. For example, the differential equation y' = x/arcsin(y). To solve this wieth separation of variables, we rewrite it as

dy / dx = x / arcsin(y)

and then separate the x and y expressions.

arcsin(y) dy = x dx

∫ arcsin(y) dy = ∫ x dx

y*arcsin(y) + sqrt(1 - y^2) = (1/2)x^2 + c

This equation cannot be solved for y in terms of x, so this expression is as simplified as it gets.

Arcsin in Algebra Problems

Inverse trig functions are used to solve problems in regular algebra or pre-calculus. For example, suppose you want to find all the values of x such that

5*sin(πx) + 7 = 2π

Solving this equation for x gives us

5*sin(πx) = 2π - 7

sin(πx) = (2π - 7)/5

πx = arcsin[(2π - 7)/5]

πx = -0.1438586

x = -0.1438586π

x = -0.0457916

This is just one value of x that solves the equation. Since the function sin(πx) is periodic with a period of 2 and is symmetric about the line x = 1/2, the other values of x that solve the equation can be found continuing the following patterns.

x = ...-3+0.0457916, -1+0.457916, 1+0.457916, 3+0.457916, 5+0.457916, ...

x = -4-0.457916, -2-0.457916, -0.45716, 2-0.457916, 4-0.457916, ...

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Comments 2 comments

Stephen Rakoczy profile image

Stephen Rakoczy 2 years ago from Pennsylvania

Thank you for doing this. It's actually not that difficult or long, it just requires some thinking.

calculus-geometry profile image

calculus-geometry 20 months ago from Germany Author

Typo fixed, thanks!

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