Download the App!
Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.
Question
Answered step-by-step
Evaluate the integral.
$ \displaystyle \int_0^{2\pi} \sin^2 \left(\frac{1}{3} \theta \right) d \theta $
Video Answer
Solved by verified expert
This problem has been solved!
Try Numerade free for 7 days
Like
Report
Official textbook answer
Video by J Hardin
Numerade Educator
This textbook answer is only visible when subscribed! Please subscribe to view the answer
Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 2
Trigonometric Integrals
Integration Techniques
Oregon State University
Harvey Mudd College
Idaho State University
Boston College
Lectures
01:53
In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.
27:53
In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.
03:16
Evaluate the integral.…
03:42
03:12
01:09
Evaluate the given trigono…
02:22
01:07
02:20
Evaluate the definite inte…
00:53
02:14
Evaluate $\int_{0}^{\frac{…
09:50
0:00
Find the definite integral…
answer the following?
this problem is from Chapter seven, Section two. Problem number eight in the book Calculus Early. Transcendental lt's a Definition by James Store. Here we have a definite overall from zero to pi. Sign square of data over three with respect that data. So for this problem, since we haven't even power on the sign and there's no co signs present, it's going to be best to use this trick in a metric identity, which says Sign square of X is one minus coastline of two x divided by two. So whatever the input is on, the sign over here in co sign multiply the same thing by two. So let's use this time to rewrite this. Integral as the integral from Sierra two pyre, they have a one half minus co sign of two times one third data. So we have ah to that over three. Also, divide this turn by, too, because we're dividing by two over here and now we can evaluate you should these integral separately. So the interval of one half with respect the data simply they divided by two. And for this integral over here we can apply you, so let's take you to be to date over three. So using this use of been evaluating this underworld the second integral we'LL tend sign of to date over three times three over too. And we're evaluating thes this expression at zero in super. So let's plug into my first for data thirteen two pi divided by two minus sign of four pi over three. And then here Let's simplify this fractions. We have three over two and then we have another two on the bottom. So let's just multiply this by three or four minus. Now we plug in Seo for data. So having zero over too, minus sign of zero over two times three over. Horrible. Did you write that? So here we have zero and from trigonometry with no sign of zero zero. So here the twos, cancel. We just have a pyre. The level minus and in sign of four pi over three is negative. Rule three and divide that by two. And then we still have three over four. So here we could just multiply these denominators together to get eight on the bottom and cancel out this double negative tub tin PIDE plus three, Route three divided by. And that's your final answer
View More Answers From This Book
Find Another Textbook