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?

First, differentiate both sides with respect to $x$:
$$0=\frac{d}{dx} \left(\int_0^x \sqrt{6+5\cos t}\;dt + \int_0^y \sin t^2\;dt\right) = \sqrt{6+5\cos x} + (\sin y^2)\frac{dy}{dx}$$
Now we have a differential equation:
$$
\sqrt{6+5\cos x} + (\sin y^2)\frac{dy}{dx} = 0.
$$
Separate variables:
$$
(\sin y^2)\;dy = -\sqrt{6+5\cos x}\;dx
$$
Now the problem is to find two antiderivatives.
