One weird integral trick

I would not have solved this one. Credit to the genius who figured out the trick. #math #maths #mathematics #shorts

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[iteerrex]

it’s one of those that you spend pages of paper and hours messing with trig identities to solve 😂

[Avighna]

For integrals, the best way to solve them is to know the answer.

[callmeandoru]

Yeah, as a university engineering student, I can say that a lot of the time, it feels like Math, Physics and Chemistry exams are less about how much you understand concepts, and more about how many weird and random tricks you know.

[angrytedtalks]

A very useful answer. I'll try to remember it. I do like a good "+ c" at the end of my integrations.

[ZzSlumberzZ]

just waiting for some Indian dude to tell me how easy this problem is

[skazzyeahh]

you can expand 4 as 4(sin^2x + cos^2x) and split the integrals then do integration by parts and you end up with -cosx/(3+4sinx)

[panda-]

You can solve with integration by parts: notice that we can match the numerator to the denominator if we write 4 = 4(sin²x + cos²x), then numerator = sin x(3+4sin x) + 4cos²x and we can split the integral into two:
∫sin x/(3+4sin x) + ∫4cos²x/(3+4sin x)². Now we can set u = 1/(3+4sin x), du = -4cos x/(3+4sin x)²dx and v = cos x, dv = -sin x dx, the integral becomes -∫u dv - ∫ v du = - vu = -cos x/(3+4sin x)

[mike.]

I had to start this video in the shorts player, then close it and go to my history, then add it to my watch later playlist, then play it from there. After watching the ad, I was actually able to watch this content in the normal player! I wish YouTube had a setting for us to tell shorts to play in the regular player, but creators can also help by just posting these as regular videos!

[Explorinity]

Dividing by cos²x is probably the first and most common trick that is taught to JEE aspirants in India.

[tylet]

Very Challenging at first but then i remembered this app called photomath…

[KM-omhm]

Actually for these type of problems there is always a tan sec relation

[kailuuu]

I just got done with my studies and about to get relax and this short recommended me

[matthewmanzanares]

The way I solved this is I sort of reversed the quotient rule for derivatives:

y = u/v
y' = (u'v - uv')/v²

I noticed that the denominator in the integral is squared therefore v = 3 + 4sinx

now all I have to do is find the value of u so that u'v - uv' is equal to the numerator inside the integral

u'(3 + 4sinx) - u(4cosx) = 3sinx + 4

I let u to be equal to cosx because of the cosine in the resulting equation


-3sinx-4sin²x-4cos²x = 3sinx+4
-3sinx-4 = 3sinx + 4

It's so close, the only difference is that the u should be negated therefore
u should be -cosx

therefore
the answer is equivalent to
-cosx/(3+4sinx) + C

and if you check the derivative you will see that it is indeed correct

[pianopanda]

Currently in pre-calc/trig and am now less confident in my math abilities haha. I’ll be there soon tho

[shrikantgupta]

1st- divide by cos²x in denomination and numerator.
2nd - then substitute 3secx+4tanx =t
then solve it it's easy.

[icelandic_mughal]

YOUR HARDEST is just the starting problem of the test.

[wooloosus]

This is definitely a problem I could've done when I was in calc BC, but definitely had me stumped now that it's been years since I had to do anything more than arithmetic lol

[darcash]

So multiply the top and bottom by something that allows the top to be the cancellation in a would be u-sub. I’m gonna create a bunch of cases for myself where this would happen. This kinda feels like almost perfect differential equations where you gotta multiply by x^m y^n or do the MyNx/N or NxMy/M trick for the integrating factor

[hellou]

When u do a lot of integrals, multiplying with sec^2/sec^2 is quite common, once u get used to it working out nicely its not such a weird trick

[rahevar]

Unfortunately in India the math text book of 12th standard which introduces to integration is filled with these kind of problems.

No matter how many rules you apply you have to do some wierd thing which you have to remember for each problem (around 100) and in exams any of those can be asked.

There also is a rule that in exams they can't ask any sum outside of the textbook so only way to get good marks is by memorizing it like its a history textbook.