PDA

View Full Version : Need help with integration


Danish_Khatri
Jul13-10, 11:08 AM
The integral of (sin(t)cos(t)) has two possible solutions: {(sint)^2}/2 and {-(cost)^2}/2 eventhough these two function are not same. Although these both solutions give same values for the definite integrals, different results are obtained when such integration appears when solving differential equations. I need some help to get out of this confusion and need to know how do we choose which solution we should use for a given problem

Gib Z
Jul13-10, 11:25 AM
Indefinite integrals can vary by an additive constant (the +C at the end). The two functions you have only vary by a constant.

HallsofIvy
Jul13-10, 04:20 PM
The integral of (sin(t)cos(t)) has two possible solutions: {(sint)^2}/2 and {-(cost)^2}/2 eventhough these two function are not same. Although these both solutions give same values for the definite integrals, different results are obtained when such integration appears when solving differential equations. I need some help to get out of this confusion and need to know how do we choose which solution we should use for a given problem
Any (indefinite) integral has an infinite number of different solutions- but they all differ by added constants. In this particular case, since sin^2(t)+ cos^2(t)= 1, cos^2(t)= 1- sin^2(t) so -cos^2(t)/2= -(1- sin^2(t))/2= -1/2+ sin^2(t)/2. Your two solutions differ by the constant "-1/2".

Danish_Khatri
Jul14-10, 12:45 PM
Thanks a lot...... The replies were really helpful