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)
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