Please!! I have to solve this for my test on Monday.
I am desperate. Please solve it step by step. Many thanks.

1. integrate x cos^2 xdx =

2. lim (1/sinx - 1/x)
(x->0)

2. *sigh* if you don't get the examples in your text, i doubt seeing a full solution here will help. there are bigger underlying problems that you need to fix, that you won't find unless you try them yourself.
Originally Posted by RinoST
Please!! I have to solve this for my test on Monday.
I am desperate. Please solve it step by step. Many thanks.

1. integrate x cos^2 xdx =
i immediately saw two ways to do this, there are others, but the easier way i saw was this:

$\int x \cos^2 x ~dx = \frac 12 \int x (1 + \cos 2x)~dx$ ....(half angle formula)

$= \frac 12 \int x ~dx + \frac 12 \int x \cos 2x ~dx$

now, the first integral is easy, for the second one, use integration by parts, with $u = x$ and $dv = \cos 2x~dx$ in the formula $\int u~dv = uv - \int v~du$

So, $\int x \cos^2 x ~dx = \frac 12 \int x~dx + \frac 12 \int x \cos 2x~dx$

$= \frac {x^2}4 + \frac 12 \left( \frac 12 x \sin 2x - \frac 12 \int \sin 2x~dx \right)$

$= \frac {x^2}4 + \frac 12 \left( \frac 12 x \sin 2x + \frac 14 \cos 2x \right) + C$

$= \frac {x^2}4 + \frac 14x \sin 2x + \frac 18 \cos 2x + C$

2. lim (1/sinx - 1/x)
(x->0)
combine the fractions to get

$\lim_{x \to 0} \frac {x - \sin x}{x \sin x} = \lim_{x \to 0} \frac {1 - \cos x}{\sin x + x \cos x}$ (Applied L'Hopital's rule)

$= \lim_{x \to 0} \frac {\sin x}{\cos x + \cos x - x \sin x}$ (applied L'Hopital's again)

$= \frac 0{1 + 1 - 0}$

$= 0$