Evaluate the definite integral. ∫ − π 4 π 4 tan y \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan y s e c 2 y d y sec^2ydy

In this problem, we're asked to evaluate the definite integral from negative pi over four to positive pi over four of tangent of y times the secant squared of y dy. Now remember that whenever we're doing a substitution on a definite integral, there are integral, there are two different ways that we could go about solving this problem. Depending on your textbook or your professor, you may have learned only one of these methods or you may just have a preference for one. In this video, I'm going to go through both methods of solving here, so feel free to skip around accordingly. The way that I'm going to start here is by using our first method we're gonna refer to as method one, where we're gonna kind of ignore those bounds until the very end, but we're gonna use our original bounds. So let's go ahead and get started with our first method here. So in method one, getting started with our steps, we wanna choose u and find du. Now with trig functions, it can be a little bit tricky to figure out what u what should be u and what does that make du. Now in this particular problem, I have the tangent of y times the secant squared of y. Now based on what we've done in other problems, you may be tempted here to make u equal to the secant squared of y. But what is the derivative of the secant squared of y? So what would du be? That's where it gets a little bit more confusing. But if instead I went ahead and chose u to be the tangent of y, I know that the derivative of the tangent is the secant squared. So if the derivative is the secant squared, that makes du secant squared of y y dy, making this work out really nicely. So when we have trig functions, we need to make sure we're thinking about what the derivative of whatever trig function we make u actually is and if it's going to make our substitution work. So here, if I choose u as tangent y that makes du secant squared y dy, which is going to work out nicely as we move into step two, rewriting our integral only in terms of this u and du. So in my integral, I can clearly see that if this is u, that makes this du. So this entire integral just becomes then the integral of u du. Now again, we're ignoring those bounds for now because we're working with this first method and we're just going to treat this as an indefinite integral at first. So now having rewritten that, I can go ahead and integrate here with respect to u. So that's going to give me here one half u squared using the power rule, and then plus my constant of integration c just for now because I'm treating this as indefinite. Now I'm gonna go ahead and replace u with our original function of x or in this case function of y. We chose u to be the tangent of y, so that's what we're gonna plug in for u here. So this becomes one half times the tangent squared of y, just writing that squared in between the way that we do for trig functions, then plus that constant of integration c. But we don't really need that constant of integration c because in this final step here, we are evaluating this at our original bounds, which in this case are from negative pi over four to positive pi over four. So getting rid of that plus c, now we're remembering that this is a definite integral. We're bounding this from negative pi over four to positive Pi over four. So going ahead and applying those bounds using the fundamental theorem of calculus here, I'm going to stick that one half on the outside. Then on the inside here, this is the tangent squared of Pi over four minus the tangent squared of negative Pi over four. Now thinking of our unit circle here, keeping that one half on the outside, what is the tangent of Pi over four? It is one. So this is one squared minus and then the tangent of negative pi over four, this is going to be negative one. So one squared minus negative one squared. Now one squared and negative one squared are both just one. So this is one half times one minus one. What is one minus one? It's just zero. So if I take zero and multiply it by one half, that is still zero. Because this works out to be zero, that means that this entire integral is just equal to zero. So that's our first method here. Let's go ahead and do this same problem using that second method where we're going to transform our bounds along with the rest of our integral when we're doing our substitution. So our method two here, starting with our same integral from negative pi over four to positive pi over four of the tangent of y times the secant squared of y. Getting started with our first step here, same exact thing we did earlier, choosing u finding du based on what I mentioned before. If u is equal to the tangent of y that then makes du equal to the secant squared of y dy which works out nicely integral. So with that done moving into our second step rewriting our integral only in terms of u and du in this method, method two, this involves actually changing our bounds as well. So with that integral we know that we have u and du having that dy in there. I forgot to write earlier secant squared y dy. This becomes du. So I know that I have the integral of u du, but what should my bounds be on my new integral? I need to go ahead and transform my bounds by plugging them into that function g of x, in this case, g of y. So g of negative pi over four and g of positive pi over four. These are going to give me my new bounds here. I'm plugging in my original bounds in order to get my new bounds based on my substitution. So I'm plugging them into the tangent of y. So this is the tangent of negative Pi over four. This gives me negative one. And then this is the tangent of positive Pi over four, which gives me positive one. So my new integral here rewriting it is going to be from negative one to positive one based on these bounds that I found here of u d u. So now we can actually move on to our third step here, integrating with respect to u. So if I integrate with respect to you here, this is going to give me a u squared or one half u squared bounded from negative one to positive one. Then I can go ahead and easily plug in those bounds. That's one half times one squared minus negative one squared. We see the same exact thing happen that we did in our first method, one squared is one, negative one squared is also one, so this is just one half times one minus one which is just zero. Since this cancels out to zero that means that our whole integral is equal to zero. And of course, we got the same exact answer that we got from using method one. Whatever method you choose to use, it should work and you should get the same answer regardless. If you have any questions, feel free to let us know, and I'll see you in the next one.

8. Definite Integrals

Substitution

Calculus

8. Definite Integrals

Substitution

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Multiple Choice

Evaluate the definite integral.
$\int_{- \frac{π}{4}}^{\frac{π}{4}} \tan y$ $s e c^{2} y d y$

0.308

$π$

$pi over 4$

Verified step by step guidance

Recognize that the integral to evaluate is $ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan(y) \sec^2(y) \, dy $. This involves the product of $ \tan(y) $ and $ \sec^2(y) $, which suggests a substitution method might be useful.

Let $ u = \tan(y) $. Then, the derivative of $ u $ with respect to $ y $ is $ \frac{du}{dy} = \sec^2(y) $, or equivalently $ du = \sec^2(y) \, dy $. This substitution simplifies the integral.

After substitution, the integral becomes $ \int u \, du $, where the limits of integration also change. When $ y = -\frac{\pi}{4} $, $ u = \tan(-\frac{\pi}{4}) = -1 $. When $ y = \frac{\pi}{4} $, $ u = \tan(\frac{\pi}{4}) = 1 $. Thus, the integral is now $ \int_{-1}^{1} u \, du $.

Evaluate the integral $ \int u \, du $ using the power rule for integration: $ \int u \, du = \frac{u^2}{2} + C $. For definite integrals, the constant of integration $ C $ is not needed. Substitute the limits of integration: $ \left[ \frac{u^2}{2} \right]_{-1}^{1} $.

Compute $ \left[ \frac{u^2}{2} \right]_{-1}^{1} $ by substituting the upper and lower limits: $ \frac{(1)^2}{2} - \frac{(-1)^2}{2} $. Notice that both terms are equal, so the result is 0. This shows that the integral evaluates to 0.