1-80 Evaluate the integral. ∫0^π/ 4 tan^5 θsec^3 θd θ |...

Name: Problem 38, Chapter 7: Calculus Early Transcendentals
Uploaded: 2021-12-10T03:51:56-08:00
Duration: 6 min 32 s
Channel: M Z
Description: 1-80 Evaluate the integral. ∫0^π/ 4 tan^5 θsec^3 θd θ

Key Concepts

Trigonometric Identities and Derivatives

A strong understanding of trigonometric identities, such as sec²? = 1 + tan²?, and the derivatives of trigonometric functions, notably d/d? (tan?) = sec²?, is essential. These identities allow one to rewrite and simplify complex trigonometric integrals, facilitating the application of the substitution method or other integration techniques.

Substitution Method

The substitution method is a powerful integration technique where a new variable is introduced to simplify the integrand. In problems involving powers of tangent and secant, setting u equal to tan? is common because its derivative, sec²?, appears in the integrand. This substitution transforms the original trigonometric integral into a polynomial or simpler function of u, making the integration process more straightforward.

Integration of Trigonometric Powers

This concept involves integrating functions where trigonometric functions like tangent and secant are raised to a power. In such cases, one often separates part of the integrand that can be easily associated with the derivative of a trigonometric function, or employs reduction formulas to simplify the expression before integrating.

Transcript

00:01 Okay, so the integral that we'd like to evaluate is the integral from 0 to pi over 4 of tangent to the 5th theta times secant cubed theta d theta.

00:15 So let's start by just looking at the indefinite integral, and then we'll use that to evaluate the definite integral.

00:21 So we want to evaluate this tangent 5th theta, the integral of tangent 5th theta times secant theta d theta.

00:29 Okay, so first off, i'm going to rewrite this in a way that might look a little funny, but it's going to make sense in a second why i'm doing this.

00:39 So i'm going to rewrite this as an integral of tangent -squared theta, and then we still got one more tangent theta times secan, cubed theta, d -theta.

00:48 So i just rewrote that.

00:51 You can make sure that's the same thing.

00:53 And then what we want to do is we want to use this trig -identity, tangent -squared theta, is equal to secant squared theta minus 1.

01:03 Okay, so now what we want to do is replace the tangent squared theta with ccant squared theta minus 1.

01:10 So what we can do is that this is secant squared theta minus 1 squared times tangent theta, secant, cubed theta, d theta.

01:20 Great, okay.

01:21 And then now we want to make a u sub.

01:26 So now if we let u equals secant theta, then du taking the derivative of both sides, we end up with the du should be secant theta times tangent theta, d theta.

01:43 Okay, so now we just wanna substitute in.

01:46 So this is equal to the integral of, so now substituting in, this is u squared minus one squared.

01:54 And then now we have this tangent, we have a copy of tangent theta times secant theta, d theta.

02:01 Well, that's going to be replaced by our du.

02:05 That's what du is equal to.

02:07 And then we still have a copy of secant squared theta left, so that's just going to be a u squared.

02:13 Awesome.

02:14 Okay, so now we can just simplify this...