I solved it a bit differently: I wrote everything in terms of cosine, then used the identity _2sin^2(x/2)=1-cos (x),_used the substitution _u = x/2 - 45°,_ arrived at _tan^2 (u)_ and then used the identity _tan^2 (u) = sec^2 (u) - 1_ to get _2tan (x/2 - 45°) - 2(x/2 - 45°) = 2tan (x/2 - 45°) - x + 90° = 2tan (x/2 - 45°) - x + C._ Your answer is way prettier than mine, though.
Alternative by using 1+sinx=(cos(x/2)+sin(x/2))^2 , same for 1-sinx, then you get integral of (tan(x/2-pi/4))^2, which can be done by converting into sec
I think I may have discovered a new (albeit, not that useful) trig identity. Specifically, another answer to this integral is 2tan(x/2-pi/4)-x+c. That makes tan(x/2-pi/4)=tan(x)-sec(x)+c, where c=0.
I think the solution set "2(tanx - secx) - x + C" does not encompass the complete collection of primitives of the integrand. For example, f(x) = 2(tanx - secx) - x + C sign(x - pi/2) is also a primitive. This is because the domain of the integrand is the union of sets (pi/2 +kpi, 5pi/2 +kpi) for all integer k, so the most general solution would be something like "2(tanx - secx) - x + sum_(k in Z) C_k * (characteristic function of (pi/2 +kpi, 5pi/2 +kpi) )", for an arbitrary collection of real constants C_k.
I am planning on doing some! Are there any specific problems you wish to see? Here is the link to 2017 qualifier problems. www.mit.edu/~same/pdf/qualifying_round_2017_test.pdf
LetsSolveMathProblems 😃 Thanks, I think 8, 10 and above all seemed hard to me, you can pick whatever is interesting to you. Thank you so much for making these videos
dunno if 9 might be too easy but could you if it´s easy do it on a side note? cause i know it´s not hard, i know the only "trap doors" are the negative sign of the cosine in the solution and the half power. but i´m kinda falling into both after another atm^^ and 13 looks like a real joyride too. (tbh i just wanna see how hyperbolic functions are dealt with^^ i mentioned, my math education is a mess^^ SO if you leave wishes open i would be dumb to not ask for the problems i think i´d learn the most^^)
I think it is safe to assume that this question was one of the easier ones
Yes, I could solve this since high school
I solved it a bit differently: I wrote everything in terms of cosine, then used the identity _2sin^2(x/2)=1-cos (x),_used the substitution _u = x/2 - 45°,_ arrived at _tan^2 (u)_ and then used the identity _tan^2 (u) = sec^2 (u) - 1_ to get _2tan (x/2 - 45°) - 2(x/2 - 45°) = 2tan (x/2 - 45°) - x + 90° = 2tan (x/2 - 45°) - x + C._ Your answer is way prettier than mine, though.
You can also substitute
t = tan(x/2) and get a rational function in t that you can integrate with partial fractions
Alternative by using 1+sinx=(cos(x/2)+sin(x/2))^2 , same for 1-sinx, then you get integral of (tan(x/2-pi/4))^2, which can be done by converting into sec
I think I may have discovered a new (albeit, not that useful) trig identity.
Specifically, another answer to this integral is 2tan(x/2-pi/4)-x+c. That makes tan(x/2-pi/4)=tan(x)-sec(x)+c, where c=0.
Elemental Lithium that is not something new I'm afraid.
This identity was present in our textbook.
Not a new identity. This identity has been widely known for millennia.
Actually I also got pretty much the same ( -2cot(x/2 + pi/4) - x + constant)
loool how arrogant are you?
Nice!
substitute (x = pi/2 - u) and write every thing in terms of cosine and then make use of the tan half angle formula
you can write the second step directly as (secx -tanx)^2 and skip a few steps/identities.
any idea on integrating that?
how can integrate that form??
Such an easy one ! Don't think , they will give this type of questions ! :)
I think the solution set "2(tanx - secx) - x + C" does not encompass the complete collection of primitives of the integrand. For example, f(x) = 2(tanx - secx) - x + C sign(x - pi/2) is also a primitive. This is because the domain of the integrand is the union of sets (pi/2 +kpi, 5pi/2 +kpi) for all integer k, so the most general solution would be something like "2(tanx - secx) - x + sum_(k in Z) C_k * (characteristic function of (pi/2 +kpi, 5pi/2 +kpi) )", for an arbitrary collection of real constants C_k.
you could simplify sec^2(x)-2sec(x)tan(x)+tan^2(x) to (sec(x)-tan(x))^2 ^^
that's what i thought too
hard to integrate
You have any idea how to integrate (sec(x) - tan(x))^2?
le Goushque Yes, actually. Not difficult, just perform a substitution.
Just a waste of time :)
Better not to simplify *_*
Hi! Great video, this is a very straight forward solution.
I have a question. Where did you studied and what (or currently studying)?
Can u do 2017 mit bee qualifiers?
I am planning on doing some! Are there any specific problems you wish to see? Here is the link to 2017 qualifier problems. www.mit.edu/~same/pdf/qualifying_round_2017_test.pdf
LetsSolveMathProblems 😃 Thanks, I think 8, 10 and above all seemed hard to me, you can pick whatever is interesting to you. Thank you so much for making these videos
dunno if 9 might be too easy but could you if it´s easy do it on a side note? cause i know it´s not hard, i know the only "trap doors" are the negative sign of the cosine in the solution and the half power. but i´m kinda falling into both after another atm^^ and 13 looks like a real joyride too. (tbh i just wanna see how hyperbolic functions are dealt with^^ i mentioned, my math education is a mess^^ SO if you leave wishes open i would be dumb to not ask for the problems i think i´d learn the most^^)
Thanks Sir
Thank you
Can you integrate this (e^x^2)(x^2)
i dont believe this function has an elementary answer
Sir I did it by substituting
X = ( π/2 + Z )
Math_mega Nilmadhab explain more plz
I simply this integral I got -(2cosx/(1+sinx) +X)+C...is right sir
Sir more easy method is replace sinx by cos(90-x)
Akhilesh Katiyar How is that any better?
nice
This one is very easy.
Easy, this us taught in highschools
What's your point?
@@Pritzelita why is a highschool level question in the MIT integration bee? i think thats his point.
Sir this is very simple integration
Easiest ever encountered