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Verify by differentiation that the formula is correct.

$ \displaystyle \int \cos^2 x \,dx = \frac{1}{2}x + \frac{1}{4}\sin 2x+ C $

$\frac{2 \cos ^{2} x}{2}=\cos ^{2} x$

00:49

Frank L.

Amrita B.

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Campbell University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

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All right. What they want you to do is take the derivative of the right side Of this equation is 1/2 X plus 1/4 Sign of two x plus e. Um And if the derivative, what I should happen is I get it equal to the left side. So my whole goal is to work through this until I get to cosign squared of X. That's our goal. So as you work through this, the derivative of 1/2 x is just one half. And then this piece is the chain rule where the derivative of sine is co sign. You leave the two X alone, but then you have to multiply by the derivative of two X. Which would be to uh And then the directive of a constant is zero. So the only thing really to fix is that this too. And uh um This 1/4 would reduce to 1/2. Uh So what I would do next is look up a double angle formula. Um And I guess I'm assuming that you could either look this up or it's easily accessible. Um That co sign of two x. I'm just gonna look pretty quick. Uh Double angle formula And we can find one as a co sign of two. X. Is equal to cosa To co sign squared of X -1. So how about I do this in green that this is equal to co sign squared of x minus one. So now if I distribute this one half in here, what I have is one half plus. Well the two half of two is one that's coastline square two X minus one half. And that's the real connection. Is that the one half minus one african cell. And I get it equal to this. And once I get equal to this I've proven my my drift. It was correct. So I don't I really shouldn't just circle this, but all of this is my work to explain that the answer was correct that the director of the anti director was correct.

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