Given information:

The function are f(x)=∫0sinx1+t2dt and g(y)=∫3yf(x)dx

Concept used:

The fundamental theorem of calculus, part 1 is defined by,

If f is continuous on [a,b] , then the function g defined by,

  g(x)=∫axf(t)dt

  a≤x≤b is an antiderivative of f , that is g′(x)=f(x) for a<x<b .

Calculation:

Using Part 1 of the Fundamental Theorem of Calculus, the derivative of the function is obtained as:

  f(x)=∫0sinx1+t2dt

By, evaluation theorem,

  ddx∫0sinx1+t2dt=F(sinx)F(x)=f′(x)=1+sin2x

Now, for g(y)=∫3yf(x)dx ,

  g(y)=∫3y∫0sinx1+t2dtdx

Using Part 1 of the Fundamental Theorem of Calculus,

  g′(y)=∫3yF(x)dx [since, 1+t2 is continuous and 0<x<b ]

Again, Using Part 1 of the Fundamental Theorem of Calculus,

  g″(y)=F(y)g″(y)=1+sin2yg″(π6)=1+sin2(π6)g″(π6)=1+(12)2g″(π6)=1+14g″(π6)=54g″(π6)=52

Therefore,

The value of g″(π6) is 52 .