different trig identities different integrals

[DuctTapePro]

i tried integrating this in 2 different way

1/(1 + x²)

first is substituting the x to tan ⦵
second is x to cot ⦵

∫ [1/(1 + tan² ⦵)] (sec² ⦵) d⦵ = arctan x + C

∫ [1/(1 + cot² ⦵)] (- csc² ⦵) d⦵ = - arccot x + C

they yield different answers :/

 

[Deleted member 4993]

i tried integrating this in 2 different way

1/(1 + x²)

first is substituting the x to tan ⦵
second is x to cot ⦵

∫ [1/(1 + tan² ⦵)] (sec² ⦵) d⦵ = arctan x + C

∫ [1/(1 + cot² ⦵)] (- csc² ⦵) d⦵ = - arccot x + C

they yield different answers :/

Not really:

tan^-1(x) + cot^-1(x) = (pi)/2

 

[Dr.Peterson]

i tried integrating this in 2 different way

1/(1 + x²)

first is substituting the x to tan ⦵
second is x to cot ⦵

∫ [1/(1 + tan² ⦵)] (sec² ⦵) d⦵ = arctan x + C

∫ [1/(1 + cot² ⦵)] (- csc² ⦵) d⦵ = - arccot x + C

they yield different answers :/

No, they're the same answer -- with different C's! Do you understand how Khan showed this?

You might want to try graphing the two answers (picking any C you like) in order to see that it is true.

This is a common stumbling block for students, so it's great that you are trying different solution methods and discovering this for yourself.