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Show that the integral ∫(1-2 sin^2⁡x)/(1+2sinxcosx) dx = (1/2) ln2 between the limits π/4 and 0. [5 marks]

First, we use the trig identities: cos2x=cos^2x-sin^2x, cos^2x+sin^2x=1 and sin2x=2sinxcosx to transform the integral to ∫(cos2x)/(1+sin2x)dx.
We know that ∫f'(x)/f(x)dx = ln|f(x)|+c, so we let f(x)=...

AC
Answered by Abby C. Maths tutor
15644 Views

Write (√(18)+10)/√(2) in the form: p + q√2 [4 marks]

First, simplify √18 by writing it as √(9x2) = √9 x √2 = 3√2.
Then, rationalise the denominator by multiplying both top and bottom of the fraction by √2.
This gives √2 x √2 = 2 in the denominator...

AC
Answered by Abby C. Maths tutor
20357 Views

f(x) = 4x^2 + 8x - 5 ; complete the square to find the turning point of f(x).

4x^2 + 8x - 5 = 0 4(x^2 + 2x - (5/4)) = 0 4((x + 1)^2 - 1^2 - (5/4)) = 0 4((x + 1)^2 - (9/4)) = 0 4(x + 1)^2 - 9 = 0 4(x + 1)^2 - 9 = 0 Turning point is at (-1, -9)

CK
Answered by Charlotte K. Maths tutor
9473 Views

Solve the simultaneous equations: y=x^2+4x-2, y=x+2

Equations: 1). y=x^2+4x-2 2.)y=x+2 x+2=x^2+4x-2 First put equation 1 in equation 2 0=x^2+3x-4 Make one side of the equation equal to 0 0=(x+4)(x-1) Factorise the equations Equations 3).x+4=0 4). x-1=0 One...

PA
Answered by Phoebe A. Maths tutor
3275 Views

(4-2x)/(2x+1)(x+1)(x+3) = A/(2x+1)+B/(x+1)+C(x+3) Find the values of the constants A, B and C

First, multiply throughout by the denominator of the main function to give as follows: 4-2x = A(x+1)(x+3) + B(2x+1)(x+3) + C(2x+1)(x+1) Then, choose values of x which will cause two of the constants to va...

MC
Answered by Michael C. Maths tutor
14238 Views

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