By expressing cos(2x) in terms of cos(x) find the exact value of the integral of cos(2x)/cos^2(x) between the bounds pi/4 and pi/3.
cos(2x)=cos2(x)-sin2(x)=2cos2(x)-1
Therefore:cos(2x)/cos2(x)=(2cos2(x)-1)/cos2(x)=2cos2(x)/cos2(x) - 1/cos2(x)=2 - 1/cos2(x)=2 - sec2(x)
Integral of sec2(x) = tan(x)
Integral of 2 = 2x
[2x - tan(x)] between pi/4 and pi/3
= (2pi/3 - tan(pi/3)) -(pi/2 - tan(pi/4))
= (2pi/3 - sqrt(3)) - (pi/2 - 1)
= pi/6 - sqrt(3) + 1
HF
Related Maths A Level answers
All answers ▸show that y = (kx^2-1)/(kx^2+1) has exactly one stationary point when k is non-zero.
Let R denote the region bounded by the curve y=x^3 and the lines x=0 and x=4. Find the volume generated when R is rotated 360 degrees about the x axis.
When solving a trigonometric equation, like sin(x) = -1/3 for 0 ≤ x < 2π, why do I get an answer outside the range? Why are there many correct answers for the value of x?
Given that dx/dt = (1+2x)*4e^(-2t) and x = 1/2 when t = 0, show that ln[2/(1+2x)] = 8[1 - e^(-2t)]