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Numerade Educator



Problem 52 Hard Difficulty

Make a substitution to express the integrand as a rational function and then evaluate the integral.

$ \displaystyle \int \frac{\cosh t}{\sinh^2 t + \sinh^4 t} $


$-\frac{1}{\sinh t}-\tan ^{-1}(\sinh t)+C$


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Video Transcript

let's make a substitution to rewrite this immigrant as a rational function. Since I see this hyperbolic co sign in the numerator, I suggest that I should take you to be hyperbolic sign. Then do you is hyperbolic co sign so we can rewrite this integral on top. There should be a dt there. Then we just write that is to you that takes care of the hyperbolic co sign in the DT Because of this equation over here and then on the bottom, that's just you. Square puts you to the fourth. So let's write that as it may be right, this is one over. We could pull out a use square and then we get you squared, plus one left over that plot that do you now we can do partial fraction the composition here. So first we should check that this factor right here is not reducible. So we have to look at the discriminatory B squared minus four a. C. Here, B is the coefficient in front of the U term you to the first power. There is no such term here, so that means be a zero for ism for is there and then a is the number that's in front of the U squared. Here's just one and C is also the constant which is one. This is a negative number that tells us that the quadratic will not factor. So here, on the other hand this is you times you So this is what the author calls case too. This is when you have a repeatedly near factor. So for the use cleared I'LL get over you be over you squared But then for the use square plus one using what the author calls Case three we have seen you plenty overyou squared plus one So let's just go ahead and keep the integral there And then these two terms that will circle on Blore equal. All we did was the partial fraction to composition and now we actually want to go ahead and see what a, B, C and D are so looking at the circle expressions here let's just multiply both of them by this denominator on the left. So the left hand side, I just get left over a one and then I have a use square plus one, and now so should have a you there in the air So it really put you out here. Another you b U squared, plus one and then see you plus D And then times you squared squad and multiplied his own a cube. Hey, you! Do you swear d seeing cute, Do you swear? And finally, let's combine like term. So let me pull out of you cute Plus, he plus u squared, Give us the pull out. You were just left with the A and in the constant storm leftovers of B. So looking at the left pee inside. There's no you Cube. There's no use square. There's no U that tells us that a plus in it b plus C and a roll zero. So there we also get that CIA zero using the first equation B plus the equal zero And then the last term on the right hand side is the constant that must equal. The constants are among the left. That means he is one. And then using this equation here, along with B equals one, we get d equals negative one. So we found all four values ABC, Andy licious plug him into this previous integral and started reading. But I'm running out of room so I'll need a new page here. So let's go ahead and plug in those values we have won over use weird and then minus one over you squared plus one to you. Now we can go ahead and just use the power will for the first Integral. This is just negative one over you. And for the second one, this is a very common in a rule. This is just work ten of you, and make a comment about that. If you forgot that fact, then you could go ahead and do it Tricks up here u equals Santa. Does she work? So go ahead and do that, if you must. The last up here, this is just used the definition of you sold me The real definition of you're not different tricks up. So here in the previous page, we chose you to be hyperbolic sign. So now we just replace Look and then we have minus arc tan and then you, which is again signed h hyperbolic sine, plus our constancy of immigration. And that's your final answer