What is #f(x) = int x^2sinx^3 + 6cosx dx# if #f(pi/12)=-8 #?

1 Answer
Ananda Dasgupta
Feb 26, 2018

#f(x) = -1/3 cos((pi/12)^3)+6 sin (pi/12)-9.22#

Explanation:

#f(x) = int x^2 sin(x^3) dx+ int 6 cos x dx#
#= 1/3 int sin(x^3) times 3x^2 dx +6 sin x #
# = 1/3 int sin(x^3) d(x^3) +6 sin x #
# = -1/3 cos(x^3)+6 sin x +C#

We can find #C# from the given condition #f(pi/12)=-8#

#-8 = f(pi/12) = -1/3 cos((pi/12)^3)+6 sin (pi/12) +C#

so

#C = 1/3 cos((pi/12)^3)-6 sin (pi/12)-8 ~~-9.22#

and thus
#f(x) = -1/3 cos((pi/12)^3)+6 sin (pi/12)-9.22#

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