Ok, need help with integrating some trigo functions
sin5x cos x dx (between pi/3 and 0)
6 cos 2x cos x dx (pi/2 and -pi/2)
Got stuck for both questions
Next,
A curve is such that d2y/dx2 = 6x -2. The gradient of the curve
at the point (2,-9) is 3
Express y in terms of x
Show that the gradient
of the curve is never less than -16/3
This is very urgent. Having prelims in 1 days time, so
please help!
For integration at O level, you do not have product rule (and
for trigo you can only integrate sin cos power of 1 and sec ^2
only). Thus you will have to rewrite the expression in a suitable
form before integrating.
1)Using the fact that sin A + sin B = 2 sin[(A+B)/2]cos
[(A-B)/2]
Let sin(5x)cos(x) = 0.5(sin A + sin B)
(A+B)/2 = 5x
A+B = 10x
(A-B)/2 = x
A-B = 2x
A = 6x and B = 4x
Thus sin(5x)cos(x) can be rewritten as 0.5sin6x + 0.5sin4x
Therefore int[ sin(5x)cos(x)]
= int [ 0.5sin(6x) + 0.5sin(4x)]
= [(0.5)([-cos 6x]/6 + (0.5)[-cos 4x]/4] your limits
= your answer
2) Same concept as (1). Use the appropriate factor formula.
3)dy/dx = 3x^2- 2x +c
At (2,-9), dy/dx = 3
3 = 3(4)-2(2)+c
c= -5
Therefore dy/dx = Gradient(G) = 3x^2-2x-5
Min/Max of anything occur when d(anything)/d(whatever) = 0
The "anything" we are interested in is the gradient (G).
Thus max/min of G will occur when dG/dx = 0
dG/dx = 6x -2
When 6x-2 = 0
x = 1/3
When x = 1/3, G = 3(1/3)^2-2(1/3)-5 = -16/3
d^2G/dx^2 = 6 (min)
Minimum value of G is -16/3.
Therefore blah blah blah(shown)
oppsss... forgot to to find y for you but you can right? Just
integrate dy/dx and sub in (2,-9) to solve the constant.
adjusted some typos as well.
