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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.$\displaystyle \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{1 + \sin x - \cos x}\ dx$

## $\ln (1+\sqrt{3})-\ln 2 \approx 0.3119$

#### Topics

Integration Techniques

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

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### Video Transcript

Let's use the formulas from exercise fifty nine. We took he, by definition to be tension affects over, too. And we use this to prove the other three equations. Let's quote and use that to evaluate this in a roll over here. So the first thing is, when we go ahead and take ecstasy over three, that's going to change the lower limit of integration here. So let's plug this in for X. See those tension of pi over six and that becomes route three. So that's our new lower limit. Similarly, playing the upper limit for X and then this becomes T equals changing a pirate for which is one that's our new upper limit. Then here we have a DX atop. So we used the ex formula over here two over one plus he square up top with the DC send in the bottom one plus, and then we have signed minus co sign. So we used these over here so sine minus coz I that will be two t trust he squared minus one all over one plus he square just adding, er subtracting here This minus co sign leave our numerator residence and then on the denominator, we can get a common denominator of one plus t swear. So let's pull that out and then in the top will have one plus T's where Plus two Tea plus T Square and I'm running out of room here. Somebody had to simplify, so we most upsides happened bottom of this one. Bye. One plus he squared. So when we do so, we should obtain Tootie Square plus two tea. And that's for the numerator within the dominant. Now cancel those one plus T square terms. We could also cancel those twos. She's the one left over three one. I went up top and then T square plus tea, and we can even pull out a team there. T plus one. I'm running out of room here. Let me go to the next page. Let's look at that in a grand one over's He t plus one. That's just one over T r a. O ver. T sees me be over t plus one and then solving for just multiplying both sides by this. Then I'm leader on the left. Go ahead and pull out a t three, and then we see that because there's not here on the left A plus B must be zero. The constant term on the left is a constant term on the right, and the left is one on the right. It's a so we must have a equals one. And then using this equation here, you could be his negative one. So then are integral becomes from roots, Rita one one over tea, one of the t plus one. Now we can evaluate that. If the second in a rolls bothering you because of that plus one here, feel free to do it yourself in the end points on route three and one. So now we just go ahead and plug those in Ellen one minus Ellen too, and then minus. And then here. So this is from plugging in won. And now we plug in the Route three and now we go ahead and cancel that that zero. We have a plus here, So that's a ln off three plus one. And then we could go ahead and combined these other two logarithms by multiplying them together and pulling out that mine design. And that's a final answer

JH

#### Topics

Integration Techniques

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp