integral of [sin(lnx) ]/ x ² using substitution

.∫ [sin(lnx) ]/ x ² dx

use the substitution lnx= u or x = e^u

then use the std. formula for integral of {e^(ax)} sinbx given here

or

use integration by parts and then manipulation as in this example

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If x+18y+c=0 is a normal to y = 5x ² -12x + 1 , find the value of c

let x+18y+c=0 be a normal at (h,k) on the curve y = 5x ² -12x + 1

implies k = 5h ² - 12h+ 1 -------------(1)

y = 5x ² -12x + 1
diff. w.r.t.x dy/dx = 10x-12

slope of normal at (h,k) = -1 / (dy/dx) = -1 / (10h-12)
also slope of the given line x+18y+c=0 is -1/18

therefore -1 / (10h-12) = -1/18

solving 10h-12 = 18 or h =3
using equation(1) k = 5*3 ² - 12*3+ 1 = 10

therefore h=3 , k=10

x+18y+c=0 is a normal at (3,10) and should be satisfied by (3,10)

3 + 18*10 +c =0

c= -183






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