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).
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)]
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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