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Numerade Educator



Problem 61 Hard Difficulty

Use the substitution in Exercise 59 to transform the integrand into a rational function of t and then evaluate the integral.

$ \displaystyle \int \frac{1}{3 \sin x - 4 \cos x}\ dx $


$$\frac{1}{5} \ln \left|\frac{2 \tan (x / 2)-1}{\tan (x / 2)+2}\right|+C$$


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Video Transcript

let's use exercise fifty nine to transform this into grand to a rational function, hopefully should be easier to compute. So from fifty nine, these were the formula. So we started off with this definition of tea, and from there we proved the remaining formulas. So these are all given Scott and plugged these all in to the original integral. So up top, we just have the X. So use this formula here one. So that's the numerator, Then on bottom, we have to re sign. So that's just three times to tea. This is over one. Plus, he swears that starts in t up there. Let me backtrack one plus he square and then minus four times co sign. And so that's four and then we have one minus T square. Now we can write this. Combine those terms on the denominator. So we have a T Square, plus one of them denominator and a top. We have sixty, but then we also see a plus foresee, and then a minus four. So this is in the bottom. And then let's go ahead and cancel those denominators of t squared plus one and then we have a part of me here. That should have been a four t squared. Then we have four C square plus. It's T minus four. It's good. And divide top and bottom here by two. Some DDT on top to T Square three T minus two. Now we can go ahead and look at that. The nominee Let's go to the next patient are running out of room. It's couldn't factor that denominator that'LL be two t minus one T plus two and now we do partial function one over two T minus one T plus two. This is a over two T minus one and then be over T plus two. So this is one equals a T plus two B two T minus one, and we could rewrite this Freddie hand side by factoring out of tea. We get a plus two B and then to a minus B on the left. We see that there's no chief, so we must have that a plus two b, a zero on the left. The constant term is one. So on the right, we must have to A minus. B is one. So's pretties equations together and soft as equals, one that here equals zero. So if let's attract two times the second equation from the first two ways cancel, then we have negative B minus four B. It's minus five. B equals one Sobhi his negative of fifth. And then here we know that a equals negative to be so eh equals two cents. So now that we have our values for early and be, let's go to the next patient right out that in a girl it was two over five and then he was negative. One over five Now for this first integral and might help you here to do you, sir, If that too, in the minus one is bothering you. And similarly, over here you could take you to be t plus two. So when we evaluate, we get to over five. And then we also get a one half from this use of institution Ellen two t minus one, and that for the second inaugural minus one, number five, it's a five on the bottom natural log T plus two. And now we should go ahead. And at that constancy of integration. So now we could maybe go ahead and clean this up a little bit. These twos will cancel so we could not take out one over five. Then here we have a log minus a log so we can write this as a fraction. That's a minus. And then t close to plus e. So here we're just using the fact that Ellen, eh, minus Ellen be is Ellen ever be. And then finally, we can use the fact that from Problem fifty nine we have that he it's tangent of X over two solutions. Bloody plug that in for tea and then we have to tangent That's over too minus one tangent affects over to And then this time plus two. And then there's our constancy, and that's the final answer.