What is #f(x) = int -cos6x -3tanx dx# if #f(pi)=-1 #?

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Answer 1 · 2016-09-08T08:15:23

Answer is:

#f(x)=-1/6sin(6x)+3ln|cosx|-1#

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

#f(x)=-intcos(6x)dx-3inttanxdx#

For the first integral:

#6x=u#

#(d(6x))/(dx)=(du)/dx#

#6=(du)/dx#

#dx=(du)/6#

Therefore:

#f(x)=-intcosu(du)/6-3intsinx/cosxdx#

#f(x)=-1/6intcosudu-3int((-cosx)')/cosxdx#

#f(x)=-1/6intcosudu+3int((cosx)')/cosxdx#

#f(x)=-1/6sinu+3ln|cosx|+c#

#f(x)=-1/6sin(6x)+3ln|cosx|+c#

Since #f(π)=-1#

#f(π)=-1/6sin(6π)+3ln|cosπ|+c#

#-1=-1/6*0+3ln|-1|+c#

#-1=3ln1+c#

#c=-1#

Therefore:

#f(x)=-1/6sin(6x)+3ln|cosx|-1#