4 ii) 2010
As we know that sin 3x sin x= 0.5 (cos 2x- cos 4x) from the previous part of the question, we can substitute this into our given integral:
integral sign (0.5 cos 2x- 0.5 cos 4x)
if we integrate this, we get
(sin2x)/4 - (sin4x)/8
substituting limits, we get
((sin (2pi/3))/4 - (sin(4pi/3))/8) - ((sin (2pi/6))/4 - (sin(4pi/6))/8)
=> (sqrt 3)/8 - (-(sqrt3)/16) - (sqrt 3)/8 - (-(sqrt 3)/16)
=> 2(sqrt 3)/16
=>(sqrt 3)/8
question 10 2009
i) To find x at M, we must find the maximum of the curve, which is simply the derivative of the curve equated to zero.
y=x^2*(sqrt(1-x^2))
we find the derivative by using the product rule,
dy/dx= 2x*(sqrt(1-x^2))+0.5*(-2x)/(sqrt(1-x^2))*x^2
= 2x(sqrt(1-x^2)) - x^3/(sqrt(1-x^2))
dy/dx=0
2x(1-x^2)/(sqrt(1-x^2)) - x^3/(sqrt(1-x^2))=0 ......... i've equated the denominators if u were unsure
(2x- 3x^3)/(sqrt(1-x^2))=0
Now we only require the numerator to be zero. Therefore,
2x-3x^3=0
x(2-3x^2)=0
Hence, x=0 and 2-3x^2=0
=> 3x^2=2
=> x^2=2/3
=> x= sqrt(2/3)
therefore, the x coordinate of M is sqrt(2/3). It cannot be 0, as M is obviously not centered at x=0
ii)As we are integrating by substituting, we must also change the variables around a bit:
x=sin(theta)
dx/d(theta)=cos(theta)
dx=cos(theta) d(theta)
To do this, we merely substitute in sin (theta) and the new variables to replace dx into the equation given:
Area=integral sign sin^2(theta)*sqrt(1-sin^2(theta))*cos(theta) d(theta)
Area=integral sign sin^2(theta)*sqrt(cos^2(theta))*cos(theta) d(theta)
Area=integral sign sin^2(theta)*cos(theta)*cos(theta) d(theta)
Area=integral sign 0.25*(sin2x)^2 d(theta)
the limits are given by substituting into the equation x=sin(theta)
we are finding theta, so we use theta=sin^-1 x
iii)Do not think that because they give you an equation, that you can use it straight away. We cannot integrate most trigonometric integrals if they are squared, such as this. So we convert it to a form we CAN integrate, namely, a cos4x form:
0.25*(sin^2(2theta))
cos(4theta)= 1-2*(sin^2(2theta))
thus,
2*(sin^2(2theta))=1-cos(4theta)
and so
0.25*(sin^2(2theta))=(1-cos(4theta))/8
Hence we are integrating this, NOT 0.25*(sin^2(2theta))
(1-cos(4theta))/8= 1/8 - (cos(4theta))/8
So, we integrate this:
Area= integral sign 1/8 - cos(4theta))/8 d(theta)
Area= 1/8*(theta) - (sin(4theta))/32 with limits pi/2 and 0
Area= 1/8*(pi/2) - (sin(2pi))/32 - 1/8*(0) + (sin(0))/32
Area=pi/16
Hope this wasn't too confusing =P
It is however up to the standard of working you require at CIE A-level so it shouldn't be tooo hard
Ask any questions if you don't get certain parts of it!
