# Calculus: trig functions anti-derivatives

## 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
Mentor

## 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.