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)
*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.i immediately saw two ways to do this, there are others, but the easier way i saw was this:
$\displaystyle \int x \cos^2 x ~dx = \frac 12 \int x (1 + \cos 2x)~dx$ ....(half angle formula)
$\displaystyle = \frac 12 \int x ~dx + \frac 12 \int x \cos 2x ~dx$
now, the first integral is easy, for the second one, use integration by parts, with $\displaystyle u = x$ and $\displaystyle dv = \cos 2x~dx$ in the formula $\displaystyle \int u~dv = uv - \int v~du$
So, $\displaystyle \int x \cos^2 x ~dx = \frac 12 \int x~dx + \frac 12 \int x \cos 2x~dx$
$\displaystyle = \frac {x^2}4 + \frac 12 \left( \frac 12 x \sin 2x - \frac 12 \int \sin 2x~dx \right)$
$\displaystyle = \frac {x^2}4 + \frac 12 \left( \frac 12 x \sin 2x + \frac 14 \cos 2x \right) + C$
$\displaystyle = \frac {x^2}4 + \frac 14x \sin 2x + \frac 18 \cos 2x + C$
combine the fractions to get2. lim (1/sinx - 1/x)
(x->0)
$\displaystyle \lim_{x \to 0} \frac {x - \sin x}{x \sin x} = \lim_{x \to 0} \frac {1 - \cos x}{\sin x + x \cos x}$ (Applied L'Hopital's rule)
$\displaystyle = \lim_{x \to 0} \frac {\sin x}{\cos x + \cos x - x \sin x}$ (applied L'Hopital's again)
$\displaystyle = \frac 0{1 + 1 - 0}$
$\displaystyle = 0$