The area under #f(x) = cscx-xsinx# and the #x# axis over the interval #[pi/6, (5pi)/8]# is obtained by compute the integral of #f(x)#,
#F(x) = int_(pi/6)^((5pi)/8) f(x) dx = int_(pi/6)^((5pi)/8) (cscx - xsinx) dx#
Apply the Sum Rule:
#F(x) = int_(pi/6)^((5pi)/8) (cscx) dx - int_(pi/6)^((5pi)/8)(xsinx) dx#
Let
#F_1(x) = int_(pi/6)^((5pi)/8) (cscx) dx# ============> (1)
#F_2 = int_(pi/6)^((5pi)/8)(xsinx) dx# =============> (2)
Let's integrate #color(red)((1))# applying integral substitution:
Let #u= tan(x/2);# then #csc(x) = \frac{1+u^2}{2u} #
and #dx=\frac{2}{1+u^2}du #
#I_1(u)=\int \frac{1+u^2}{2u}\frac{2}{1+u^2}du = \int \frac{1}{u}du#
#I_1(u)= |ln(u)|;# substituting #u= tan(x/2)#
#F_1 = [|ln(tan(x/2))|]#
Now let integrate #color(blue)((2))# applying integration by parts. Recall from the product rule of derivative of 2 function f and g
#(f*g)'= f'g+g'f# integrate both side and you end with
#int(f*g)' dx= int(f'g+g'f) dx #
#(f*g) = int(f'g+g'f) dx# we can rewrite this as:
#color(purple)(int(g'f )dx = (f*g) - int(f'g) dx)# ======>(3)
From what we derived in (3) and letting:
#u = x; (du)/dx = u' = 1 #
#v' = sinx; v = int sinx dx = -cosx #
#color(purple)(int uv'dx = uv - int u'v dx) #
#intxsinx dx = x(-cosx) - int-cosx dx#
#F_2(x) = -xcosx + sinx = [sinx - xcosx]#
#F(x) = F_1(x) - F_2 =[lntan(frac(x)(2)) -sinx+xcosx]_(pi/6)^((5pi)/8)#
