Trig sub, then sub trig

For this trig integral, one substitution isn't enough!
There is another way to solve this integral without partial fractions by splitting up the numerator as 1/2(sin(t)+cos(t))−1/2(sin(t)−cos(t)). Then you can do a u-sub on the left part, and the right part is just a constant so that's easy as well.
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[@puneetporwal]

Instead of making the second substitution we could try multiplying and dividing by (sin theta + cos theta). These reduces the integral to a very manageable function: -0.5*(tan 2theta + sec 2theta + 1)

[@mathematix-rodcast]

That was my means of attack too. I knew the double trig functions would be a good means of attack.

[@seifaboud9501]

The cos(theta)/sin(theta)-cos(theta) can also be evaluated to Ax + Bln|sin(theta) - cos(theta), where A and B are non-zero rational numbers... In this case both A and B will be 1/2 if you split the integral into two halves and then add/subtract ½sin(theta).

[@techlover2000]

Alternatively, after the first sin substitution you could create two integrals, C which is the the original integral, and S which is the integral with the same denominator as the original but with a sin on top instead. Then C-S and C+S can both be directly evaluated and then solve for C using Gaussian elimination

[@vcvartak7111]

Once you get cos(t)/(sin(t)-cos(t)) for integration Call it I(1) then take I(2) as sin(t)/(sin(t)-cos(t)) now just do I(1)+i(2) and i(1)-i(2) and integrate both one is just the derivative in numerator of its denominator and other is just integration of dt. Then add i1(t) +i2(t) and i1(t)-i2(t) and you get 2i(t) which is needed intgration

[@josht7238]

Great explanation!! Thanks

[@adandap]

Very nice. You can however stick with the trig functions if you want. There's one idea elsewhere in the comments, but I took (I'll use x because it's easier to type, but it's your theta) 1/(cosx - sinx) = (cosx + sinx) /(cos^2x - sin^2x) = - (cosx + sinx)/cos 2x. Then in the numerator cos^2 x = 1/2 (1 + cos 2x) and sinx cosx = 1/2 sin 2x Then you just have standard integrals of double angle trig functions.

[@violintegral]

Lol I just came back to this video 3 months later and found an even easier way to evaluate the integral. Just substitute x=sin(u) right off the bat, now you have the integral of cos(u)/(sin(u)-cos(u)). Then, use a method of undetermined coefficients to break up this integral into a linear combination of integrals with known antiderivatives. We know that (sin(u)-cos(u))/(sin(u)-cos(u))=1 (provided that sin(u)-cos(u)≠0), and that the antiderivative of (cos(u)+sin(u))/(sin(u)-cos(u)) is simply ln|sin(u)-cos(u)|, since we have an integral of the form du/u. Using this knowledge, we can break up the cos(u) in the numerator into a linear combination of (sin(u)-cos(u)) and (cos(u)+sin(u)). To do this, set A(sin(u)-cos(u))+B(cos(u)+sin(u))=cos(u), and then solve for A and B by creating a system of equations using the values of sin and cos and 0 and π/2, respectively. We find that A=-1/2 and B=1/2. After rewriting the integrand, we can evaluate the antiderivative, which evaluates to 1/2ln|sin(u)-cos(u)| - 1/2u + C. Then, reversing our substitution by using the Pythagorean identity or a reference triangle, the final antiderivative is given as 1/2ln|x - √(1-x²)| -1/2arcsin(x) + C. This method of undetermined coefficients is really useful for integrals of the form (Asin(u) + Bcos(u))/(Csin(u) + Dcos(u)). This is sort of unintuitive, but once you've seen enough trig integrals, you start to learn the tricks.

[@violintegral]

I rationalized the denominator by multiplying the numerator and denominator by the conjugate of the denominator and got the integral of (x+√(1-x²))/(2x²-1), which I split using the linearity of the integral. Next, I used the substitution x=sin(theta). Once I was in the theta world, I had the integral of cos²x/(2sin²x-1), which I rewrote as -1/2(cos(2x)+1)/cos(2x), which I thought was pretty clever (but I'm too lazy to write in the thetas). Then I had to deal with getting rid of the thetas when I got -1/4ln(sec(2theta)+tan(2theta)), which was a bit difficult, but not too bad if you know your double angle formulas. I got 1/2ln(2x²-1)-1/4ln(2x√(1-x²)+1)-1/2arcsin(x) + C as my final answer, which is equivalent to yours.

[@Aek1961]

Εγώ λύω το ολοκληρωμα ως εξής πιο γρηγορα : Οταν φτασουμε στο συνθ/ ημθ- συνθ τότε πολλαπλασιαζουμε αριθμητή και παρονμαστη με το συνθ+ημθ. Τότε θα εχουμε ολοκλ. του -(συν^2θ+ ημθσυνθ)/ συν2θ= ολοκλ. του -1/2×(1+συν2θ+ημ2θ)/συν2θ= ολοκλ. του -1/2×(εφ2θ+τεμ2θ+1)= -1/4 ln|συν2θ|-1/4ln|τεμ2θ+εφ2θ|-1/2θ+c= ........................+c

[@ahmadm.l.suleyman..5310]

Another approach: We multiply the numerator and denominator by ( x + sqrt(1-x^2)) we'll get : Integral( -(x+sqrt(1-x^2)) dx ) and then we substitute (x) with sin(u) and we'll get: Integral ( -(sin(u)+cos(u))*cos(u)) du) I think this integral is so easy to make.

[@MuPrimeMath]

Be careful: the denominator doesn't become 1 when you do that! See www.wolframalpha.com/input/?i=Expand+%28x-sqrt%281-x%5E2%29%29%28x%2Bsqrt%281-x%5E2%29%29

[@ahmadm.l.suleyman..5310]

@@MuPrimeMath I've made an unforgivable mistake. I apologize to mathematics 😓

[@gabrieleciccarello4876]

That's awesome!

[@holyshit922]

I prefer Euler substitution sqrt(1-x^2)=xu-1

[@viktor-kolyadenko]

А что делать после 3:12 нас на матане преподаватель очень хорошо обучил.

[@muneebahmad7729]

Nice video.... 🖤