Відео

Where can taken coaching for putnam preparation

Nice

It is interesting to note that an alternative solution, using Feynman’s method, has almost the same ingredients, though Laplace is less “ad hoc” than Feynman.

which language is this?

hello sir remember me?

@5:45 1/2

great

We , when preparing for jee , do such problems for boosting confidence. 😅

Υπάρχει πολύ πιο εύκολος τρόπος: απλώς γράφουμε το e^x= 1/e^-x. Κάνουμε ομώνυμα στον παρονομαστή και έχουμε τελικά: e^-x . sinx / 2+e^-x cosx +e^-x sinx H παράγωγος του παρονομαστή είναι -2e^-x . sinx άρα -1/2 . ln|2+e^-x(sinx+cosx)|+c

Thank youu teacher❤

Another metods, us with the hyperbolic funcions, i dont know why, but the solucion can be 1/2 arctgh(x^2) Anyway, thanks for the classic method :D

i learned integration 2 days Ago And When This Came I was Like Eh This Was MIT question 😂😂 So easy Proceeds to get fked Up

another way, substitute logx=t so x=e^t, and dx=e^t * dt, so integral (e^t)^(1/t) * e^t * dt, which is integral e * e^t dt on solving is e*e^t + c on substituting e^t as x final answer is e*x + c.

thank you soo much for this lecture! what a nice shortcut 🥳

Hello sir. Can you find me the Laplace transform of integration sinx/x from 0 to 1 . Thank you

I did it without using gamma or Beta functions ua-cam.com/video/wKALjs60lBs/v-deo.html

This was asked in kcet 2023 lol

ua-cam.com/video/yPvy0XQjeng/v-deo.html

Excellent sir!

Compute the Integral of x^2024/(1+x^2)^2023 please.

Sir deserve million subscribers

Beautiful short and crisp lectures sir. Loved it

Es muy bueno.

Just some feedback, great video but can you please improve your handwriting (mine sucks too so i know the struggle). I thought the Gamma was a square root, and I thought the square in the denominator was a γ, euler-mascheroni.

Great work ❤

Prof they look scary 😮

Thank you so much sir .....

Thank you very much 🫡

Thank you for this , sir !

can be done easily ..separate the numerator into two and use his method .ans is done within 2 min .

We can also try this by substituting x equal to 1/t

Amazing keep the good work👌👌👌

nice

Your 2 looks like r if we have r instead of 2 we need to split it into cases

Superb! Thank you!

Very easy

As Jee Aspirant , like these questions are too easy 🗿🗿

The first integral is easily written as \int dx\ x \frac{d}{dx}\ \exp{ln^2x} which upon integration by parts, yields x\exp(ln^2x} - \int dx\ \exp{ln^2x}. The second term cancels against \int dx\ \exp{ln^x} in the original integral, leaving you with x \exp{ln^2x} + c.

Easiest integral ever

Sir, plz check whether you made a small mistake of not putting negative sign on 2ln[1/3^(1/2)] and -6/3^(1/2)at the last line of the solution!

Hello sir

Lot of mistakes

e^log x = x therefore (e^log x)^1/log x = e = x^1/log x

would love to see you solve gaussian integral

answer is wrong because you did not consider left and right hand limit. Square root of x^2 is not x it is |x| so, its is 2/3 sin|x|/x which is 1 for 0+ and -1 for 0- therefore limit does not exist

You forgot to put the absolute value at the end, otherwise you may get a negative result if you decide to evaluate the integral.

I have subscribed to your channel

Nice explanation sir

Beautiful sum sir !!!

Sir good explaiantion try to do more qnss