TheLivingMartyr Posted September 2, 2012 Share Posted September 2, 2012 If two expressions are exactly equivalent, then are their integrals exactly equivalent? I was trying to work out, without using integration by parts (trying to avoid infinite series and all that) [math] \int sin(x) cos(x) dx [/math] So naturally, I consulted my double angle formulae, and saw that [math] sin(2x) = 2 sin(x) cos(x) [/math] which obviously implies that, [math] sin(x) cos(x) = \tfrac{1}{2} sin(2x) [/math] The integral of the RHS is an easy one, so I just did, [math] \frac{1}{2} \int sin(2x) dx = -\tfrac{1}{4} cos(2x) + c [/math] and so assumed that, [math] \int sin(x) cos(x) dx = -\tfrac{1}{4} cos(2x) + c [/math] but I then check Wikipedia, and a couple of integral calculators for good measure, and they tell me the actual integral is [math] -\tfrac{1}{2} cos^2(x) [/math] and since [math] \tfrac{1}{4} cos(2x) =! \tfrac{1}{2} cos^2(x) [/math] I'm now a bit stumped as to why my integral is wrong. All the below are confirmed to be correct, [math] sin(2x) = 2 sin(x) cos(x) [/math] [math] \int sin(2x) = -\tfrac{1}{2} cos(2x) [/math] by the same sources which told me the integrals were different! For God's sake, you can even go on one of those graph plotters and ask it to plot the integrals of sin(x)*cos(x) and (1/2)*(sin(2x)), and it plots the same graph twice!!! I'm tearing my hair out here, can somebody please tell me if I'm just missing something obvious, or if some of my sources are incorrect? Link to comment Share on other sites More sharing options...

sws5000 Posted September 2, 2012 Share Posted September 2, 2012 your answer is correct too they are the same because 1- let u=sin(x) then integral= (1/2)*[sin(x)]^2 2- let u = cos (x) then integral =-(1/2)*[cos(x)]^2 sum tow integrals =( 1/2)*{[sin(x)]^2-[cos(x)]^2}=-(1/2)*cos(2x)...... ( this for tow integrals ,but we have one integral then we must devide on 1/2) and got finally =-(1/4)*[cos(2x)] it means you and they are correct dont wory i wrote in this way because i think you will understand what i wanted to say if you dont understand i will write more clear for you your answer is correct too they are the same because 1- let u=sin(x) then integral= (1/2)*[sin(x)]^2 2- let u = cos (x) then integral =-(1/2)*[cos(x)]^2 sum tow integrals =( 1/2)*{[sin(x)]^2-[cos(x)]^2}=-(1/2)*cos(2x)...... ( this for tow integrals ,but we have one integral then we must devide on 1/2) and got finally =-(1/4)*[cos(2x)] it means you and they are correct dont wory Link to comment Share on other sites More sharing options...

ajb Posted September 2, 2012 Share Posted September 2, 2012 You have indefinite integrals only up to a constant. In particular [math]- \frac{1}{4} \cos(2x) = - \frac{1}{4}\left( 2 \cos^{2}x -1 \right)[/math] from the double angle formula. So you get the same answer up to a constant. Link to comment Share on other sites More sharing options...

TheLivingMartyr Posted September 2, 2012 Author Share Posted September 2, 2012 So we have three possible integrals for this function, sin(x)*cos(x).... which are [math] \frac{sin^2(x)}{2} [/math] [math] \frac{-cos^2(x)}{2} [/math] [math] \frac{-cos(2x)}{4} [/math] and these expressions satisfy the following equalities [math] \frac{1}{2} - \frac{cos^2(x)}{2} = \frac{sin^2(x)}{2} [/math] [math] \frac{-cos(2x)}{4} - \frac{1}{4} = \frac{cos^2(x)}{2} [/math] [math] \frac{sin^2(x)}{2} - \frac{1}{4} = \frac{-cos(2x)}{4} [/math] .....Is this something to do with the constant of integration? and if so, then how do you know which one to use when calculating a definite integral? Sorry ajb, my reply was an inb4, it just took ages to type Link to comment Share on other sites More sharing options...

ajb Posted September 2, 2012 Share Posted September 2, 2012 Sorry ajb, my reply was an inb4, it just took ages to type No problem. .....Is this something to do with the constant of integration? and if so, then how do you know which one to use when calculating a definite integral? It will not matter which one you use, the constants will simple cancel when calculating a definite integral. Link to comment Share on other sites More sharing options...

TheLivingMartyr Posted September 2, 2012 Author Share Posted September 2, 2012 oh damnit, on my 5th line of TeX, the RHS should be negative, sorry! OK, thanks everyone! Wasn't sure if I'd made a mistake or whatever, thanks for clearing that up! Link to comment Share on other sites More sharing options...

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