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Use the substitution in Exercise 59 to transform the integrand into a rational function of t and then evaluate the integral.
$ \displaystyle \int_0^{\frac{\pi}{2}} \frac{\sin 2x}{2 + \cos x}\ dx $
$$4 \ln \frac{2}{3}+2$$
Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 4
Integration of Rational Functions by Partial Fractions
Integration Techniques
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Let's use the substitution from exercise 59. That's this equation over here. And these were the results of that exercise. We ended approving these equations here based on the substitution. So using that substitution, let's go ahead and rewrite this in overall. So first I see that I can rewrite this as to sign times Go sign. So let me just go ahead and pull out that too, and then let me instead of writing sign, go ahead and use our right hand side over here. So that's two t atop So that sign and then for co sign we have another equation here. So that's our numerator. We have two times sine times co sign and we make that integral longer for t X. It's just this over here. So this is all in our numerator and then oops, there should be a d. T. At the end here too. Uh huh. And then on the bottom two plus co sign. So two plus one minus t squared over one plus t squared. And because we're in a new variable teeth, we should go ahead and change those limits of integration. So how to do that? Just plug in these original limits into X in this equation over here and you'll get your T limits so the lower limit will be t equals tangent of 0/2. So it's J zero for the upper limit. And then if we do pi over two over two, that's power for that's one. So there's an upper limit. Let's go ahead and simplify this. Multiply out all these tools up here. Get an eight. Pull that out. So this is our numerator. After we multiply all those quadratic together, we have a third power. That's the numerator the denominator. Go ahead and add those fractions together on bottom combine like terms. So this is our denominator and then go ahead and cancel. So here we could cancel this with one of these. That's where the two is coming from. Over here. Now I'm running out of room here. Let me go to the next page. So we have eight and then go ahead here and let's right. This is yeah. So this is just for the partial fraction decomposition for the previous rational function that we just had on page one. Yeah, So now there's three inaugurals for the first one here. Just go ahead and do a U sub and then for the second, very similar. And what should we do for the last one? Same as the original one. Mm. Okay, so here's the first integral eight. But then we also have that, uh, minus. Here. Let's pull out that minus. And we have a two here to divide by after the use of you could change those limits of one and two for the next inaugural. Eight in the two again, this time from 3 to 41 over you, do you? And then for the last 1 8/2, 1 to 2. Do you ever u squared and we know how to evaluate. All these here is negative for l N 1 to 2 for Ellen three and four. And then we have minus four over you from 1 to 2. Let's go ahead to the next page and plug those in. So we have negative for l n two plus four Ln four minus four l n three and then minus two plus four. And this could finally be written as two plus four Ln for over six, and then maybe go ahead and just cancel. Simplify that fraction, and that's our final answer.
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