Calculus: trig functions anti-derivatives

[Melawrghk]

Homework Statement

Let g'(x)=-17x¹⁶* sin(Ax⁹) - 9Ax²⁵*cos(Ax⁹)

g(1)=143/9
Where A is a real number such that tanA=1/sqrt(80), 0<A<pi/2

Find g(0)

The Attempt at a Solution

I was able to get the antiderivative of g'(x), so that:
g(x) = -x¹⁷*sin(Ax⁹)

I got A=6.3794 degrees.

BUT, I don't get g(1)=143/9

What am I doing wrong? Did I assume wrong that the g'(x) was created by the product rule?

 

[Mark44]

Homework Statement

Let g'(x)=-17x¹⁶* sin(Ax⁹) - 9Ax²⁵*cos(Ax⁹)

g(1)=143/9
Where A is a real number such that tanA=1/sqrt(80), 0<A<pi/2

Find g(0)

The Attempt at a Solution

I was able to get the antiderivative of g'(x), so that:
g(x) = -x¹⁷*sin(Ax⁹)

I got A=6.3794 degrees.

BUT, I don't get g(1)=143/9

What am I doing wrong? Did I assume wrong that the g'(x) was created by the product rule?

Your antiderivative looks fine. Did you remember to add a constant? IOW, you should have
g(x) = -x¹⁷*sin(Ax⁹) + C

You're sort of given A, and you're given that g(1) = 143/9, so you can find C. If you did all that, you can still run into problems calculating things with a calculator, such as when it's in radian mode but you're working with degrees.