Calculate the derivate $dx/dy$ using $\int_0^x \sqrt{6+5\cos t} \, dt + \int_0^y \sin t^2 \, dt = 0$ I want to calculate $\frac{dx}{dy}$ using the equation below.
$$\int_0^x \sqrt{6+5\cos t}\;dt + \int_0^y \sin t^2\;dt = 0$$
I don't even know from where to start. Well I think that I could first find the integrals and then try to find the derivative. The problem with this approach is that I cannot find the result of the first integral.
Can someone give me a hand here?

HINT: You have $$f(x)=\int_0^x\sqrt{6+5\cos t}\,dt=-\int_0^y\sin t^2 \,dt=g(y)\;.$$ What are $\dfrac{df}{dx}$ and $\dfrac{dg}{dy}$ according to the fundamental theorem? And when you have $\dfrac{df}{dx}$, what can you multiply it by to get $\dfrac{df}{dy}$?
