Evaluate the integrals in Exercises 23-32 . ∫0^π/ 2 θ√(1-cos2 θ) d θ

Name: Problem 31, Chapter 8: Calculus
Uploaded: 2020-09-01T20:11:43-08:00
Duration: 4 min 56 s
Channel: Prakash Hampole
Description: Evaluate the integrals in Exercises 23-32 . ∫0^π/ 2 θ√(1-cos2 θ) d θ

Evaluate the integrals in Exercises 23-32 . ∫0^π/ 2...

Key Concepts

Definite Integration

Definite integration involves evaluating an integral over a specified interval. This process not only requires finding an antiderivative but also properly applying the limits of integration, which sometimes involves additional adjustments if a substitution method has been used.

Integration of Trigonometric Functions

This concept focuses on the techniques and strategies used specifically for integrating functions that involve trigonometric terms. Integrals of trigonometric functions often require the use of identities, substitutions, and careful consideration of the periodic and symmetric properties of those functions to simplify and accurately evaluate the integral.

Trigonometric Identities

Trigonometric identities simplify expressions within integrals by rewriting functions in equivalent forms. For instance, identities like the double-angle formulas can transform complex trigonometric expressions into more manageable ones, which is essential when the integrand contains functions such as cosine or sine in non-standard forms.

Substitution Method

The substitution method is a technique used to simplify an integral by changing the variable of integration, thereby transforming the integrand into a form that is easier to integrate. This is particularly useful when an integrand includes composite functions or when a trigonometric expression calls for a change of variables to match a standard integration form.

Transcript

00:01 Before we find this definite integral, we should find this indefinite integral.

00:06 And we often want to look at, when we look at these integrals, you want to think about any identities we already know.

00:13 One that you may remember is cos 2 theta is equal to, it is equal to 1 minus 2 sine squared theta.

00:31 And if we enter that in here in place of goes 2 theta, we can cancel some things out.

00:40 We have 1 minus 1 minus 2 sine squared theta, and when you subtract, you get inside the square root 2 sine squared theta.

01:08 Now, when you square with that, you get integral theta times square root of 2 times square root of sine squared, which is absolute value sine theta.

01:20 We don't really like to have absolute values, but we don't really need one for this because if we just look at the interval that it's being integrated along, which is 0 to pi over 2.

01:38 Well for that sign is always positive so this is just integral of theta we can take the square root outside and sign theta because that's the same as absolute value theta for this range interval and from here we have some we if we differentiated this we would just eliminate the variable and we also know how to integrate this sign so we can do integration by parts.

02:15 We will let u be theta and therefore d u equals d theta and our d v will be sign theta d theta.

02:27 So by integration v will be negative cose theta and therefore this by integration by parts is equal to uv so negative theta v -du, so negative cose theta, d -u, or d -theta.

02:56 No, at the end, we're not gonna put an arbitrary constant because we will be finding a definite integral anyway.

03:04 And we have this, and the integral of, well, these two negatives cancel out, so we can just do this, and the integral of cos is sine theta.

03:21 So we don't really like having negatives first, so we'll do this.

03:33 And there is our indefinite integral...

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