Thanks guys for the help provided.
Hello, tester85!
I'll explain the first one for you.
It's quite challenging and uses a technique you may not have seen.
Integrate by parts . . .$\displaystyle I \;=\;\int e^{ax}\sin bx\,dx$
. . $\displaystyle \begin{array}{ccccccc}u &=& \sin bx & & dv &=& e^{ax}dx \\ du &=& b\cos bx\,dx & & v &=&\frac{1}{a}e^{ax} \end{array}$
We have: .$\displaystyle I \;=\;\tfrac{1}{a}\,e^{ax}\sin bx - \tfrac{b}{a}\int\! e^{ax}\cos bx\,dx $
Integrate by parts (again):
. . $\displaystyle \begin{array}{ccccccc}u &=&\cos bx & & dv &=& e^{ax}dx \\ du &=& \text{-}b\sin bx\,dx & & v &=& \frac{1}{a}e^{ax} \end{array}$
We have: .$\displaystyle I \;=\;\tfrac{1}{a}\,e^{ax}\sin bx - \tfrac{b}{a}\bigg[\tfrac{1}{a}e^{ax}\cos bx + \tfrac{b}{a}\int\! e^{ax}\sin bx\,dx\bigg] + C$
. . $\displaystyle I \;=\;\tfrac{1}{a},e^{ax}\sin bx - \tfrac{b}{a^2}\,e^{ax}\cos bx - \tfrac{b^2}{a^2}\underbrace{\int\! e^{ax}\sin bx\,dx}_{\text{This is }I} + C $
. . $\displaystyle I \;=\;\tfrac{1}{a}\,e^{ax}\sin bx - \tfrac{b}{a^2}\,e^{ax}\cos bx - \tfrac{b^2}{a^2}\,I + C $
Multiply by $\displaystyle a^2\!:\;\;a^2I \;=\;ae^{ax}\sin bx - be^{ax}\cos bx - b^2I + C $
. . . . . . . .$\displaystyle a^2I + b^2I \;=\;ae^{ax}\sin bx - be^{ax}\cos bx + C $
Factor: .$\displaystyle (a^2+b^2)I \;=\;e^{ax}(a\sin bx - b\cos bx) + C $
. . . . . . . . . . .$\displaystyle I \;=\;\frac{e^{ax}}{a^2+b^2}(a\sin bx - b\cos bx) + C $
Therefore: .$\displaystyle \int e^{ax}\sin bx\,dx \;=\;\frac{e^{ax}}{a^2+b^2}(a\sin bx - b\cos bx) + C$
alternatively you can use the well known identity:
$\displaystyle
\sin bx = \frac{{e^{jbx} - e^{ - jbx} }}
{{2j}}$
so the integral becomes a simple combination of exponents:
$\displaystyle
\int {e^{ax} \sin bxdx = \frac{1}
{{2j}}\int {e^{x(a + jb)} - e^{x(a - jb)} dx} }
$
$\displaystyle \frac{1}
{{2j}}\int {e^{x(a + jb)} - e^{x(a - jb)} dx} = \frac{1}
{{2j}}\left[ {\frac{1}
{{a + jb}}e^{x(a + jb)} - \frac{1}
{{a - jb}}e^{x(a - jb)} } \right] =
$
$\displaystyle
= \frac{1}
{{2j}}e^{ax} \left[ {\frac{{ae^{jbx} - jbe^{jbx} - ae^{ - jbx} - jbe^{ - jbx} }}
{{a^2 + b^2 }}} \right] = \frac{{e^{ax} }} {{a^2 + b^2 }}\left( {a\sin bx - b\cos bx} \right)
$
$\displaystyle \int{\frac{x^2}{(x-1)(x+1)^2}\,dx}$
Use partial fractions.
$\displaystyle \frac{x^2}{(x-1)(x+1)^2} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$
$\displaystyle = \frac{A(x+1)^2 + B(x-1)(x+1) + C(x-1)}{(x-1)(x+1)^2}$
Since the denominators are the same, the numerators are equal...
So $\displaystyle x^2 = A(x+1)^2 + B(x-1)(x+1) + C(x-1)$
Expand and you get
$\displaystyle x^2 + 0x + 0 = (A + B)x^2 + (2A + C)x + A - B - C$
Equating like powers of x gives
$\displaystyle A + B = 1, 2A + C = 0, A - B - C = 0$.
Solving the equations simultaneously gives
$\displaystyle A = \frac{1}{4}, B = \frac{3}{4}, C = -\frac{1}{2}$.
So $\displaystyle \frac{x^2}{(x-1)(x+1)^2} = \frac{1}{4} \left(\frac{1}{x-1}\right) + \frac{3}{4} \left(\frac{1}{x+1}\right) - \frac{1}{2} \left(\frac{1}{(x+1)^2}\right)$
$\displaystyle \int{\frac{x^2}{(x-1)(x+1)^2}\,dx} = \frac{1}{4} \int{\frac{1}{x-1}\,dx} + \frac{3}{4} \int{\frac{1}{x+1}\,dx} - \frac{1}{2} \int{\frac{1}{(x+1)^2}\,dx}$
$\displaystyle = \frac{1}{4} \ln |x-1| + \frac{3}{4} \ln |x + 1| + \frac{1}{2(x+1)} + c$
For the last one, you need to set up the partial fractions as
$\displaystyle \frac{A}{x+1} + \frac{Bx+C}{x^2 + 1}$.
You should find $\displaystyle A = \frac{1}{2}, B = -\frac{1}{2}, C = \frac{1}{2}$.
Knowledge that $\displaystyle \frac{-x + 1}{x^2 + 1} = \frac{-x}{x^2 + 1} + \frac{1}{x^2 + 1}$ should also help you out.
I'd like to help, but I'm not sure if my interpretation is correct. Do you mean
$\displaystyle \int_0^{2\pi}\sqrt{\frac{1-\cos x}{2}}\,dx$ or
$\displaystyle \int_0^{2\pi}\sqrt{1-\frac{\cos x}{2}}\,dx$ or
$\displaystyle \int_0^{2\pi}\sqrt{1-\cos \left(\frac{x}{2}\right)}\,dx$??
--Chris