Understanding the Two Solutions for Integrating (sin(t)cos(t))

[Danish_Khatri]

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]

Indefinite integrals can vary by an additive constant (the +C at the end). The two functions you have only vary by a constant.

 

[HallsofIvy]

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]

Thanks a lot... The replies were really helpful