What is #f(x) = int -cos6x -3tanx+cot(x/3) dx# if #f(pi)=-2 #?

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Answer 1 · 2016-05-27T04:47:10

#f(x) = int( -cos6x -3tanx+cot(x/3)) dx #

# => f(x) = -intcos6xdx -3inttanxdx+intcot(x/3) dx#

# => f(x) = -1/6sin6x -3ln|secx|+3ln|sin(x/3)| +c ...color(red)"(1)"#

,where c= integration constant

Now imposing the given condition #f(pi)=-2#

# -1/6sin(6xxpi) -3ln|secpi|+3ln|sin(pi/3)| +c=-2#

#1/6xx0-3xx0+3xxln(sqrt3/2)+c=-2#

#c=-2-3ln(sqrt3/2)#

Inserting the value of c in # equation color(red)"(1)"# we have

# f(x) = -1/6sin6x -3ln|secx|+3ln|sin(x/3)| -2-3ln(sqrt3/2)#