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Can't we use the direct pattern for such a question? Integration dx/root(x^2-a^2) = log | x + root(x^2-a^2) | + C

sinx - cosx = sqr(2). sin(x - pi/4) and NOT sqr(2). sin(pi/4 - x)

Simplemente aplica la fórmula para esa integral indefinida y al poner los límites de integración colocar con límite en x=1, lo cual se evalua directamente y el resultado es rápido y sencillo.

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The problem looks very intimidating at first glance!

couldnt you just let the bottom be 0.5sin2x

honestly mind blowing

Thank you so much for taking the time to make another video based on my comment! I truly appreciate the effort, and your explanation has helped me a lot. Keep up the amazing work! And I actually messed up with the calculation,thats why it didn't show the correct result

It looks like that interference pattern from the double slit experiment

I have put sin2θ=2tanθ/(1+tan²θ) formula,but its showing different result

15:56 isn't du/dx = -4/x^5?

too easy

Just use DI method

who would have guessed the sin square factor simply halves the gaussian integral. interesting video keep it up

Why such a long solution? Can’t you directly apply kings rule. Convert cot to tan, take lcm and add? You will get answer within 15 seconds.

sin(x)+1 = cos^2(x/2)+2cos(x/2)sin(x/2)+sin^(x/2) sin(x)+1 = (cos(x/2) + sin(x/2))^2 sin(x)+1 =(sqrt(2)*(cos(x/2)*1/sqrt(2) + sin(x/2)*1/sqrt(2)))^2 sin(x)+1 =(sqrt(2)*cos(x/2 + pi/4))^2 sqrt(sin(x)+1) = |sqrt(2)*cos(x/2 - pi/4)| sqrt(sin(x)+1) = sqrt(2)*|cos(x/2 - pi/4)| sqrt(2)*cos(x/2 - pi/4) Indefinite integral = 2*sqrt(2)*sin(x/2 - pi/4)+C Indefinite integral = 2*sqrt(2)*sin(x/2 - pi/4)*sqrt(2)*cos(x/2 - pi/4)/sqrt(1+sin(x))+C Indefinite integral = 2*2*sin(x/2 - pi/4)*cos(x/2 - pi/4)/sqrt(1+sin(x))+C Indefinite integral = 2*sin(x - pi/2)/sqrt(1+sin(x))+C Indefinite integral = -2*cos(x)/sqrt(1+sin(x))+C

Why does e^x become e^2x?

It's a very lengthy method Just divide the function inside integration by x^2 and Use 1+1/x^2 = t Solve and get ln√2 as answer

Please quit the music, it's painful in headphone, raw voice is the best

I prefer the original version without background music it makes us focus more on the problem

simple and concise, great video!

Love from India ♥ I read in class 12 and similar problem came to my exam one week back

im really new to integration, but integrating e to the power e to the power 2016x separated wouldn't have been easier?

Coshx why epower x -epower-x

oh man, the 6048x is added to the power

too ez for 12th grade

x^4 + x^2 + 1 = (x^2 + x + 1) (x^2 - x + 1) Also (x^2 + x + 1) - ( x^2 - x +1) = 2x Putting this in Numerator and then will have two term with a Quadratic in Denominator which if easily Integratable

Integral e^x (f(x) + f'(x)) dx = e^x f(x) + c

You can easily apply rule : integration of 1/x^2+a^2 = 1/a tan^-1 x/a. Don't even have to assume thita after u

this is a pretty good Q

Whenever there is too much of log in integration just put lnx = u and x=e^u, then dx = e^udu ez

Got the answer in my head in a few seconds. The way I thought about it was recalling that ∫f'(x)f(x)dx = 1/2[f(x)]² + c. From experience, having a log(x) and a 1/x in the same integral means a reverse chain rule with a function in the form log(f(x)) is involved (an example being ∫ logx/x dx), so I just thought about what if y = log(log(x)), and so dy/dx = d/dx(log(x)) / log(x) = 1/xlog(x), using y = log(f(x)) ⇒ dy/dx = f'(x)/f(x).

Yup I got it and did the same thing! Thanks for the video!

Oh yes you did it at the end, nice! my bad

U dont need to do all of that, Its just ln|cosh(x)| by using reverse chain rule

That's definitely one way to do it, but I did it a different way where I saw that e^(e^x+e^-x+x) = e^(e^x + e^-x) * e^x Likewise, e^(e^x+e^-x-x) = e^(e^x + e^-x) * e^-x You can then pull out the e^(e^x + e^-x) to get (e^(e^x + e^-x)) * (e^x - e^-x) After this it's just a simple u sub, u = e^x + e^-x, du = e^x - e^-x dx so the integral is simply e^u, whose integral is e^u, undoing the sub gives you e^(e^x + e^-x) + C, which is equivalent to e^2coshx + C I wouldn't be surprised if both forms were treated as correct on the exam

Just use integration by parts 😂😂

seems to me that this is not too hard but my lack of experience and effort put into these questions restricts me to think stuff like this within a reasonable period of time

Hats off for you sir. It is very clear and elegant solution thank you so much

nice video

I was wondering about using integral of dx/√x²+a² => ln |x + √x²+a²| + C

It could be much easier if you extract the common factor “e to the power of (ex + e minusx )”at first and the answer is basically the derivative of itself

Don't you think it would be better to substitute log(logx) as u? I reckon in that case the integral can be solved in less number of steps.

Too much working. Instead: tan²x = sec²x - 1 sec²x = 1/cos²x Therefore, sinx[(1/cos²x) -1] = tanxsecx - sinx Integrate this = secx + cosx +c

I did with, Cosx=u

Nice video, but this only shows it is true for x < 90

I did this in high school

This would be a lot quicker in the competition because the contestants would definitely have memorized the integral of sec(x) and all the other trig functions. Great solution!

Can we see why 2(sinxcosx) is equal to sin(2x)?

Beautiful