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Express cos2x in the form a*cos^2(x) + b and hence show that the integral of cos^2(x) between 0 and pi/2 is equal to pi/a.

Apply the double angle formula to cos2x to yield the requested result. cos2x = 2cos^2(x) - 1 Spot that the question asks us to prove the value of cos^2(x) when integrated, and that we can move the variables ...
LP
Answered by Louis P. • Maths tutor
5823 Views

Solve the differential equation dx/dt = -2(x-6)^(1/2) for t in terms of x given that x = 70 when t = 0.

First, manoeuvre variables so that we can integrate the equation. 1/(x-6)^(1/2) dx = -2 dt Integrate the equation and add the constant. 2(x-6)^(1/2) = -2t +c Solve for t. t = -(x-6)^(1/2) - c Substitute x = ...
LP
Answered by Louis P. • Maths tutor
5919 Views

Complete the square of the following expression: 2x^2 -8x+21

2(x-2)^2 +13
IF
Answered by Iliya F. • Maths tutor
4117 Views

A point lies on a circles diameter such that the distance from the point to the edge of circle is 4 times the distance from the point to the centre. What is the circles area in cm^2 if the distance from the point to edge is 5cm?

first we know that the length given has a 4:1 ratio with the remaining length of the radius so 5/4 + 5 = 25/4 = radius, the formula for the area of a circle is A=pi*(r^2) so the area is 625/16*pi.
LR
Answered by Luka R. • Maths tutor
3085 Views

If y = sec(z)tan(z)/sqrt(sec(z)) then find the indefinite integral of y with respect to z.

Using the substitution u = sec(z)=> du = sec(z)tan(z) dz.So, the integral ∫ y dz = ∫ sec(z)tan(z)/sqrt(sec(z)) dz=> ∫ y dz = ∫ 1/sqrt(u) du = 2sqrt(u) + C = 2sqrt(sec(z)) + C.
JM
Answered by Jordan M. • Maths tutor
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