Question

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

Answer 1

`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`