This Problem is Smooth like Butter

take a look at this wild integral if you were in cow 2 you would probably start off with a trig substitution let's give it a try see that x squared plus 1 in the denominator that looks an awful lot like the trig identity tangent squared plus one is secant squared so if we let x equal tangent theta and substitute that into the denominator we'll get that identity we'll have to substitute tan theta in for x in the numerator and since we're changing the variable we're going to need to change the differential let's take the derivative to get dx that'll be the derivative of tangent theta which is secant squared theta d theta lastly let's just change those x bounds to u bounds if we plug in zero for x we'll see that's also zero for theta if we plug in one for x that will be pi over four for theta so we've transformed our integral but we're not quite at a place where we can integrate this yet sometimes it's easier to rewrite trig functions in terms of sine and cosine let's do that tangent is sine over cosine and let's go ahead and combine with the plus one get a common denominator we'll have sine plus cosine over cosine inside that logarithm when you have the log of a quotient it's just screaming to use properties of logarithms so let's do the log of the numerator minus the log of the denominator things seem to be cleaning up quite nicely but we're still no closer to actually integrating this we know sine squared plus cosine squared is one but what about sine to the first plus cosine to the first inside the logarithm no less this is where things get a little bit tricky notice that we're already subtracting a log of cosine maybe we can rewrite sine plus cosine in terms of just cosine essentially we're going to derive a less well-known trig identity let's just say sine plus cosine equals some constant times cosine of theta plus some other constant luckily we do have the very well-known trig identities for cosine of a sum or a difference let's just apply that here we have sine thetas on the left we have sine thetas on the right we have cosine thetas on the left we have cosine thetas on the right let's do matching coefficients the coefficients on the left side are one the coefficients on the right side well they're a little more complicated but they have to equal one now we can take advantage of the fundamental trig identity square both sides of both these equations and add them up then we can factor out the constant sine squared plus cosine squared that's one one plus one is 2. this lets us solve for one of our constants as square root of 2. substitute this value back into those two equations and take the inverse trig functions it'll let us solve for that other constant we were looking for and so whether you realize it or not sine theta plus cosine theta can be written as square root of 2 cosine theta minus pi over 4. so now once again we have the log of a product let's just split that up into separate logarithms and we'll use properties of logarithms to drop that exponent down front time for the fun part maybe you don't think we've actually done anything to help us integrate this well we could integrate that first part it's just a constant so that's not really a big deal but look at those two cosines inside the logarithm they're pretty similar one's just theta the other is theta minus pi over four and we're integrating from zero to pi over four cosine is an even function so theta minus pi over four is the same as pi over four minus theta when it's inside cosine do a u substitution here let u equal pi over four minus theta that means d u equals minus d theta and this integral can be rewritten as negative natural log cosine of u d u to change the bounds when theta is pi over 4 u will be 0 and when theta is zero you will be pi over four so look at these two definite integrals they're the same thing their underlying variables are different but if we were to compute the integrals the values would be exactly the same we have an addition of one a subtraction of the other and so they cancel each other out all that's left to do is anti-derive this constant by multiplying it by theta substitute in pi over four and there you go is our answer to that slick definite integral if you like seeing super cool integrals click the video on the screen i'll see you in that one

Channel: BriTheMathGuy

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Keywords: math, maths, putnam, putnam math, putnam mathematics, putnam exam, putnam integral, putnam integration, put nam, put nam math, put nam exam, put nam test, putnam test, put nam mathematics

Id: fxWmLyh3kWM

Length: 4min 49sec (289 seconds)

Published: Mon Jun 14 2021