Calculus II



Calculus II

MAT 146

Derivatives and Integrals Involving Inverse Trig Functions

As part of a first course in Calculus, you may or may not have learned about derivatives and integrals of inverse trigonometric functions. These notes are intended to review these concepts as we come to rely on this information in second-semester calculus.

Derivatives of Inverse Trig Functions

One way to translate into words the meaning of the function [pic] is as follows, based on right-triangle trigonometry:

“y represents the sine ratio for an angle of measure x.”

We write the inverse of this function as [pic] and translate this equation into words as

“y is the angle measure whose sine ratio is x.”

The right triangle shown here illustrates this, for the sine of angle y is opposite/hypotenuse = x/1 = x.

For use in the derivation that follows, we note that in the triangle above, ?2 + x2 = 12. Solving for the unknown, we have [pic]. From that we deduce that [pic].

Our goal is to determine the derivative of [pic]. We start that process by noting that an equivalent form for [pic] is [pic]. We now apply implicit differentiation to this equivalent form:

[pic]

We now have the derivative of the inverse sine function and, because differentiation and integration are inverse operations, we have an integral as well:

[pic]

Through a similar derivation as just presented, we can show that

[pic] and [pic]

as well as

[pic] and [pic].

Likewise, we can generate derivatives for three other inverse trig functions:

[pic]

We use these derivatives and integrals to solve additional problems involving inverse trig functions. Here are some examples, first involving derivatives and then involving integrals.

Example #1: Determine the derivative of [pic].

We use derivative rules we already know—in this case, the chain rule—as well as the new information about derivatives of inverse trig functions:

[pic]

Example 2: Determine the derivative of [pic].

[pic]

Example 3: Determine the derivative of [pic].

Here we recognize that arcsin(u) = sin-1(u). We also use the trig identity sin2(x) + cos2(x) = 1.

[pic]

Depending on the value of x, this last expression will simplify to either 1 or –1. On the interval -π  ................
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