let x = tan^2 (u) dx = 2tanu * sec^2 u du integral ( tanu * 2 sec^2 (u) * tan u /sec^2 u ) = integral of ( 2 tan^u du )= integral of ( 2sec^2 u - 2 du) = 2tanu - 2u + c= 2sqrt(x) -arctan(sqrt(x)) - c - thank you for the amazing integrals :)
Not surprisingly I prefer the digamma method, but no one has ever accused me of being normal. It would be interesting to have a competition where the objective is to solve integrals using two or three different methods of your choice per integral. It would be quite challenging, but a lot of fun!
We can also use many alternatives for this
Good point! I think there is many ways to do this one :) thanks
Beautiful! And the 3rd would of course make use of the sum of geometric series🍻
Yep good way! Thanks! 🍻🍻🍻
let x = tan^2 (u)
dx = 2tanu * sec^2 u du
integral ( tanu * 2 sec^2 (u) * tan u /sec^2 u ) =
integral of ( 2 tan^u du )=
integral of ( 2sec^2 u - 2 du) =
2tanu - 2u + c=
2sqrt(x) -arctan(sqrt(x)) - c
- thank you for the amazing integrals :)
thanks AliTea! appreciate it :)
Not surprisingly I prefer the digamma method, but no one has ever accused me of being normal. It would be interesting to have a competition where the objective is to solve integrals using two or three different methods of your choice per integral. It would be quite challenging, but a lot of fun!
Hi Mike. Good idea! Multiple Methods Integration Bee 2025