One way to get around the + C is to apply a definite integral (instead of an indefinite) on both sides when you have I'(n) = some nice expression; you can choose your upper bound as the value you want, and the lower bound as a known value on the function, then add it to the other side Applying indefinite integrals can get us some pretty formulas, however applying definite will help for speed. Also for some very tricky ones applying a definite integral is actually necessary
@andikusnadi1979Ай бұрын
Cool
@maxvangulik1988Ай бұрын
arcosh(1)=0 because (e^0+e^-0)/2=(1+1)/2=1
@vismofАй бұрын
ooo finally
@NAxxen18Ай бұрын
Hey can u make a strategy vedio which will help us to identify where we could apply the feynman trick because it's very confusing for me to where i can actually apply it
@Silver-cu5upАй бұрын
thats the hardest part about this method, I dont even know exactly for advanced cases. However, most commonly, you use this method when you come across integrals that are in the form: - ln(f(x))/f(x) -arctan(f(x))/f(x) - ln(a+g(x)) - (x^a-1)/ln(x) - (f(x))^n/ln(f(x)) These are the most common ones, especially related with logs and some arctans.