What is #f(x) = int (x^2sinx^3- cot3x)dx# if #f(pi/12)=-1 #?

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Answer 1 · 2016-02-22T23:56:57

It is

#f(x)=int -((cosx^3)')/3dx-int (log(sin3x)')/3dx=> f(x)=-1/3*cos(x^3)-1/3*log(sin3x)+c#

where c is the integration constant which we can calculate because we know that #f(pi/12)=-1#

Hence we have that

#f(x)=-1/3cos(x^3)-1/3*log(sin3x)+c=> f(pi/12)=-1/3cos(pi^3/12^3)-1/3*log(sin(3*pi/12))+c=> #

but #f(pi/12)=-1=>c=-1+1/3*cos(pi^3/12^3)+1/3log(sin(pi/4))#

Hence the function is

#f(x)=-1/3*cos(x^3)-1/3*log(sin3x)-1+1/3*cos(pi^3/12^3)+1/3log(sin(pi/4))#