conics2008 said:
this is what im trying to say
sinx cos x = 1/2 sin2x right ???
therefore 1/2 S sin2x >> 1/2 {- 1/2cos2x }
therefore Intergal of sinx cosx = -1/4 cos2x ?? right
I might be wrong, i never trust my working out on the net xD
I asked myself such question not so long ago when I was a high school student.
The integral of an integrable function is unique up to a constant.
So the above makes sense, even though
-1/4 cos2x "looks" different from 1/2 sin^2(x), The difference between these two functions is only by a constant.
since (double angle formula) cos2x = 1 - 2 sin^2(x)
hence -1/4 cos2x = -1/4 + 1/2 sin^2(x)
Which demonstrates why S sinx cox dx has two "different" primitives.
In fact, integrating trigonometric functions in two different ways can often lead to a trigonometric identity.
For examples, using the t formula to find the integral of sec x lead to one primitive function while multiplying sec x by (sec x and tan x) will yield a "different" primitive.
This phenomenon occurs thanks to the existence of many trigonometric identities.
This could lead to interesting questions to be asked in exams and textbooks however such questions rarely appear anywhere or if they exist at all.
Perhaps the above conics2008's dilemma should be written in future mathematics textbooks.
Did you know there will be a
new stage 6 mathematics syllabus to be implemented very soon.
Differential equations will be included in 4 unit and
statistics will be studied in depth in 2 unit and many other alterations. check
www.boardofstudies.nsw.edu.au and have your say. Take a look at them in your leisure time. Some of you ponder doing accelerated mathematics in year 10 may think about doing HSC later if you favour these changes.
