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Evaluate the integrals$$\int_{0}^{\pi} \sqrt{1-\cos 2 x} d x$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 8

Techniques of Integration

Section 2

Trigonometric Integrals

Integrals

Integration

Integration Techniques

Campbell University

Harvey Mudd College

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Idaho State University

Lectures

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In mathematics, integratio…

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Evaluate the integrals.

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Evaluate the definite inte…

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Evaluate the integral.

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Section eight that to number 24. We're finding the definite in a row from zero to pie square root of one minus co. Sign of two. Ex First step is I need to be able to integrate this. Find the anti derivatives. So how do I integrate square root of one Money's co sign to x d. X? I know that the sine squared of acts is going to be 1/2 one, minus the co sign two X that tells me that, too sine squared of X is equal to one minus co sign of two X. Then, if I were to take the square root of both sides of that equation, I would get the square root of two square bit of science. Squared X is equal to the square root of one minus co signed to X, which gives me the square root of two absolute value. Sign of X is equal to the square root of one minus co signed to X, so this integral turns into the square root of two and then the integral of the absolute value of the sign of X t X. So we're asked to integrate this from zero to pie. So this is the square root of two zero to pie. Absolute value Sign of X DX. Now I know that the sign of X is greater than equal to zero when X is in the interval from zero to pie, their quadrants one in two. That means that I can remove those absolute value signs. So this turns into the square root of two absolute value of sine x, d x, and so the integral of sine of excess minus coastline of X. So this is negative square root of two co sign of axe evaluated from zero to pie. And so this just becomes negative square to two co sign and pies. Negative one minus the co sign of zero Um, is one. So, um, and sort out this. There's one minus sign that I had missing here. So this is minus squared to the coastline of zeros one. So this is the square root of two minus negative square to two, which is just too square root of two. Okay, so that's the value for the definite integral

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