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Maths
A Level

Integrate y with respect to x, where y = cos(x)/[1+tan^2(x)]

y = cos(x)/sec2(x) = cos3(x)

y = cos(x)(1-sin2(x)) = cos(x) - cos(x) sin2(x)

-> sin(x) - sin3(x)/3 + c

DC
Answered by Daniel C. Maths tutor
3526 Views

Using the equation cos(a+b) = cos(a)cos(b) - sin(a)sin(b) or otherwise, show that cos(2x) = 2cos^2(x) - 1.

First let a = b = x such that:          

          cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

becomes:

          cos(x + x) = cos(x)cos(x) - sin(x)sin(x)

Leading to:

     ...

BH
Answered by Benjamin H. Maths tutor
3657 Views

A general function f(x) has the property f(-x)=-f(x). State a trigonometric function with this property and explain using the Maclaurin series expansion for this function why this property holds. Write down the integral in the limits -q to q of f(x) wrt x

Note that this property is the definition of an odd function, or draw a sketch of what this looks like in general about the horizontal axis. E.g. f(x)=sin(x) which has the expansion f(x)=x-((x^3)/3!)+((x^...

JH
Answered by James H. Maths tutor
3667 Views

Find the derivative of f(x)=x^3 sin(x)

Find the derivative of f(x)=xsin(x).

To do this calculation we need to use the product rule of differentiation: if f(x)=u(x)v(x), then the derivative is f'(x)=u'(x)v(x)+u(x)v'(x). In o...

MM
Answered by Mairi M. Maths tutor
23079 Views

Given that y=x^3 +2x^2, find dy/dx . Hence find the x-coordinates of the two points on the curve where the gradient is 4.

First of all we are asked to differentiate the function. This can be easily done by multiplying the coefficient of x by its exponent, and then decreasing the exponent by one. Therefore, dy/dx=3x^2+4x We a...

IH
Answered by Imran H. Maths tutor
15116 Views

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