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)

[erezsolomon]

At the end you can simplify the answer even more when getting a common denominator on the first natural log and then separating it to two logs. After that you write the square root as the half power and bring it outside of the natural log to get minus one over four to cancel with the plus one over four of the second term in the solution, leaving you with just two terms.

[techlover]

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

[seifaboud]

The 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).

[vcvartak]

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

[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.

[viktor-kolyadenko]

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

[holyshit]

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

[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 + C as my final answer, which is equivalent to yours.

[ahmadm.l.suleyman..]

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.

[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 (provided that sin(u)-cos(u)≠0), and that the antiderivative of 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.

[Aek]

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