Solving definite integral using a sneaky substitution

[@bokiiibrother9995]

That 2nd substitute is incredibly good ❤

[@cipherunity]

Thanks

[@Chris_387]

Nice video, may I suggest you an integral? But I have to send it as an image.

[@cipherunity]

I shall look into it.

[@qwertyman123]

what do you think of the algebra problem
also i have a fascinating number theory problem

[@cipherunity]

I can look into it

[@qwertyman123]

​@@cipherunity Let n be a positive integer and p a prime, both greater than or equal to 2. Suppose n divides p-1 and p divides n^3 -1. Prove 4p-3 is a perfect square.
n divides p-1 -> n p. But this contradicts that n =2, k>=1: we may conclude n-k-1 = 0, or k = n-1. Then 4p-3 = 4(n(n-1)+1) - 3 = 4n^2 - 4n + 4-3 = 4n^2 - 4n + 1 = (2n-1)^2. The proof is complete.

[@cipherunity]

@@qwertyman123 saved

[@qwertyman123]

@@cipherunity is the algebra problem good

[@cipherunity]

@@qwertyman123 I just saved it. But still I need to read it.

[@fengshengqin6993]

So here is the question about this substitution : How could you jump to this type ? The process in your mind is the key . Could you help to share it ?

[@cipherunity]

The 2nd substitution is very powerful. I used it in many places. It helps you to divide the logarithmic term in the numerator to suitable factors.