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Evaluate the integral.

$ \displaystyle \int_0^1 \frac{dx}{(x^2 + 1)^2} $

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$\frac{1}{4}+\frac{\pi}{8}$

03:55

Carson Merrill

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

Integration Techniques

Campbell University

University of Nottingham

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.

02:45

Evaluate the integral.…

03:38

08:35

06:20

01:13

Evaluate the integrals.

01:47

Evaluate the definite inte…

00:17

Alright, so this is an interesting definite integral because you sub doesn't work. It's not quite right for an inverse tan anti derivative or you know a derivative giving us inverse tan because it's squared. So we need another trick and one trick is just to know your trig identities. Super. Well because if you remember that tan squared of X plus one equal secret square defects. And we kind of noticed that down in there kind of has a similar form. So what we can do is we can let um X equal tan of ax or actually we're going to make it equal at the change of variable. Let's make it equal to say tan of you. So then D X. D. You equal seek and squared of you or D X. Is seek it squared of U. D. U. So we're gonna need that because we're going to substitute in. Um Actually both the trigger identity and the substitution. Alright, now when we substitute in and we have to remember to change the limits. So when X is zero then Tan has to be some angle. Well, Tan of you is zero. So we know Tana 00. So basically when X equals zero um then if Tana who is going to equal zero, you has to be equal to zero as well. So there's a zero still there. Now we're gonna do what about when X is one? Which is the upper limit, you have to think when is Tana you equal to one? That's when sine and cosine are the same or when you equals pi over four. So our integral will be from zero to pi over four. Um D. X. Will substitute as secret square view. Do you? And um notice that I can replace X. By Tana view. So 10 of U squared plus one squared. Alright. So now we can use that trig. Identity on the bottom. So on top we still have secrets squared. You do you on the bottom? Remember it can squared plus one is second squared. So we're going to get secret squared. You but the whole thing is squared Well. Right, okay. So integral pi over four integral from zero to pi over four. And I have some secret squared on top. Speaking to the fourth on the bottom, I can cancel to the secret squares. Leaving me. Uh Do you over seek it squared you? However that is um If I flip it to the top, the same as co sine squared. You can you believe this problem? So many steps. Okay then I guess what? There's another trig identity. And that trick identity is that if I have cosine squared of you, it's equal to one half one plus cosine of to you. That is crazy. Right? Look at all the steps. Um let me change color to make it more interesting. Okay, so now I can take the anti derivative. The one half goes along for the ride the anti derivative of one is you. And the anti derivative. Cosine of U. is sine of two you over to. And we're going to plug in first, pi over four. Subtract and plug in zero. When we plug in pi over four we get one half A pi over four Plus sign, let's see twice. Make it pi over two over to. Um When we plug in zero to both terms will get zero. So that's easy. So this cleans up. Well let's first figure out. Sign up. High over to that part is just one. So then if I distribute the one half, I'll get pi over eight plus one half times a half, 1/4. So that after all that cool work is a solution to our definite integral. Hopefully it all made sense. Have a wonderful day.

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