Thursday, November 20, 2014

Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 42

Determine the integral π3π6csc3xdx

Using Integration by parts

csc3xdx=udv

where


dv=csc2xdxv=cotxu=cscxdu=cscxcotxdx


then


csc3xdx=udvcsc3xdx=uvvducsc3xdx=cscxcotxcotxcscxcotxdxcsc3xdx=cscxcotxcscxcot2xdxApply Trigonometric Identity csc2x=1+cot2x for cot2xcsc3xdx=cscxcotxcscx(csc2x1)dxcsc3xdx=cscxcotx((csc3x)cscx)dx)csc3xdx=cscxcotxcsc3xdx+cscxdxCombine like terms



csc3xdx+csc3xdx=cscxcotx+cscxdx2csc3xdx=cscxcotx+cscxdxcsc3xdx=cscxcotx+cscxdx2csc3xdx=12cscxcotx+12cscxdxcsc3xdx=12cscxcotx+12(ln(cscx+cotx))+ccsc3xdx=12cscxcotx12ln(cscx+cotx)+c


Evaluating the limit from π6 to π3


π3π6csc3xdx=[12cscxcotx12ln(cscx+cotx)]π3π6π3π6csc3xdx=12csc(π3)cot(π3)12ln(csc(π3)+cot(π3))+12csc(π6)cot(π6)+12ln(csc(π6)+cot(π6))π3π6csc3xdx=1\cancel2(\cancel233)(33)12ln(233+33)+1\cancel2(\cancel2)(3)+12ln(2+3)π3π6csc3xdx=3912ln(\cancel33\cancel3)+3+12ln(2+3)π3π6csc3xdx=1312ln(3)+3+12ln(2+3)

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