Top answers

integrate (4cos^4 x -4cos^2x+1)^1/2

we must integrate (4cos 4 x-4cos 2 x+1) 1/2 factorise first ((2cos 2 x-1) 2 ) 1/2 this becomes (2cos 2 x-1)which equals cos 2xthe integral of cos 2x is1/2 sin 2x + C
DW
Answered by Dominic W. • Maths tutor
3912 Views

Differentiate (x^2)cos(3x) with respect to x

First we start off by seeing that we are multiplying together two functions both containing x, so we want to apply the product rule. As we know the product rule is (f(x)g(x))'=f(x)g'(x)+f'(x)g(x) so we can a...
AB
Answered by Arthur B. • Maths tutor
9621 Views

Write (3 + 2√5)/(7 + 3√5) in the form a + b√5

First multiply top and bottom by conjugate of denominator, (7-3√5), and expand(3 + 2√5)(7 - 3√5)/(7 + 3√5)(7 - 3√5)(21 + 14√5 - 9√5 - 30)/(49 + 3√5 - 3√5 - 45)Simplify top and botton(-9 + 5√5)/4Write in requ...
BA
Answered by Beth A. • Maths tutor
6095 Views

Expand using binomial expansion (1+6x)^3

(1+6x)^3 = 1+3(6x) +(3)(2)(36x^2)/2 + (3)(2)(1)(216x^3)/6 = 1+18x+108x^2 + 216x^3
OO
Answered by Ola O. • Maths tutor
4578 Views

Find the gradient at the point (0, ln 2) on the curve with equation e^2y = 5 − e^−x

Question is asking for gradient at x = 0, y = ln2. e^2y = 5 - e^-x. Differentiation with respect to x: 2e^2y * dy/dx = e^-x . dy/dx = e^-x / 2e^2y. At x = 0, y = ln2 ~ dy/dx = e^0 / 2e^2ln2 = 1 / 2e^ln4 = 1 ...
LK
Answered by Lokmane K. • Maths tutor
5641 Views