Trigonometry

How would I find the integral of √5cos(2x - λ) - 1 ?

$\displaystyle \int \sqrt{5}\cos (2x-\lambda) -1 .dx = \sqrt{5} \int \cos (2x-\lambda) .dx - \int 1.dx$

For the first integral, you can do it without much thinking. Though if you really must, use the sub $u=2x-\lambda$

I got (√5/2)sin(2x - λ) - x, so Im supposed to show that when I evaluate it with upper bound λ and lower bound 0, the area is 2 - λ. I cant seem to get the right answer? I thought maybe I had the wrong integral but im not sure now.

Yeah you're not getting that answer from $\displaystyle \int_0^{\lambda} \sqrt{5}\cos (2x-\lambda) -1 .dx = \sqrt{5}\sin \lambda - \lambda$

Oh right, the original equation is f(x) = 2(2cosx - sinx)sinx.
If I integrated that would It come out as 2 - λ?

Its part iv)

You need to use the fact that $\tan \lambda = 2$ and deduce the exact value of $\sin \lambda$ before substituting it into $\sqrt{5} \sin \lambda - \lambda$

EDIT: You actually find it in part (i) anyway so use it.

With the integral what happened to the 1/2 in √5/2?

Do the integral. You should end up with $-\sin(-\lambda)$ from the lower bound. Use the fact that $\sin(-x) \equiv - \sin x$ so then you just have $\sin \lambda$ from the lower bound. The half shall then cancel accordingly.