AboutAhmed Salami Expertise I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I believe i would be very helpful in calculus and can as well help a good deal in Physics with most emphasis directed towards mechanics.
Experience An engineering graduate. I have been doing maths and physics all my life.
Question can u please send information or notes that fully explains and teaches calculus for trigonometry. or maybe a site that will help me out. especially those with odd or even powers. e.g.(sin^3)or (cos^4) etc.
Answer Hi Jason,
Sorry for the time taken.
I actually don't have any links to give to you but i
could still give you some useful hints that would enable
you solve any of them by yourself easily.
You have to make use of the following identities
(cosx)^2 + (sinx)^2 = 1
cos2x = 1 - 2(sinx)^2 = 2(cosx)^2 - 1
therefore
(sinx)^2 = 1/2(1 - cos2x)
(cosx)^2 = 1/2(1 + cos2x)
These are very useful identities.
Now, you deal with odd powers by simplifying as
e.g (sinx)^3 = (sinx)^2 . sinx
= (1 - (cosx)^2).sinx
= sinx - (cosx)^2.sinx
(cosx)^3 = (cosx)^2 .cosx
= (1 - (sinx)^2).cosx
= cosx - (sinx)^2.cosx
For even powers
e.g (cosx)^4 = [(cosx)^2]^2
= [1/2(1 + cos2x)]^2
(sinx)^4 = [(sinx)^2]^2
= [1/2(1 - cos2x)]^2
To actually perform the integration, we make use of the
fact that
$[f(x)]^n . f'(x) dx = [f(x)]^(n+1) / (n + 1)
And so
$(cosx)^n .sinx dx = -(cosx)^(n+1) / (n + 1) + C
$(sinx)^n .cosx dx = (sinx)^(n+1) / (n + 1) + C
Therefore, as examples
$(sinx)^2 dx = $1/2(1 - cos2x) dx
= 1/2(x - sin2x/2) + c
= x/2 - (sin2x)/4
In the same manner,
$(cosx)^2 dx = $1/2(1 + cos2x) dx
= 1/2(x + sin2x/2) + c
= x/2 + (sin2x)/4 + c
Also,
$(sinx)^3 dx = $[sinx - (cosx)^2.sinx] dx
= -cosx + (cosx)^3/ 3 + c
Now go ahead and give some others a try and i'm sure
you'll be fine.
I hope i have helped, you can always get back to me.
Regards.
