# Calculate the derivate $dx/dy$ using $\int_0^x \sqrt{6+5\cos t}dt + \int_0^y \sin t^2dt = 0$

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?

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A $\mathrm{d}t$ went missing. Is that $\sin^2 t$ or $\sin t^2$? –  user21436 Jan 22 '12 at 21:09
I fixed it, thanks. It is $\sin t^2$. –  papas Jan 22 '12 at 21:15
Don’t bother trying to do the integrations; think about using the fundamental theorem of calculus. (You definitely won’t be able to do the second integral.) –  Brian M. Scott Jan 22 '12 at 21:17

For $\int_0^x f(t)\,dt + \int_0^y g(t)\,dt = 0$, I'd start by rewriting to $$\int_0^x f(t)\,dt = -\int_0^y g(t)\,dt = u$$ where I've introduced a new variable for the common value of the two sides. This allows us to imagine that the curve is parameterized by $u$.

The fundamental theorem of calculus then gives us $\frac{du}{dx}=f(x)$ and $\frac{du}{dy}=-g(y)$, and we can then get $\frac{dx}{du}$ and $\frac{dy}{du}$ as the reciprocals of those, and finally we can find $\frac{dx}{dy}$ by implicit differentiation. Thus we don't actually need to evaluate the two integrals!

(This is exactly the reverse procedure from finding the curve by solving the resulting differential equation by separation of the variables).

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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 $\dfrac{df}{dy}$?

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I don't consider myself dumb but I think that I won't be able to solve it by myself. –  papas Jan 22 '12 at 21:38
@papas: Take a look at Michael Hardy’s answer, but don’t try to find antiderivatives: just rearrange the last expression to get $dx/dy$ on one side. You’ll still have both $x$ and $y$ on the other side, but I don’t think that you can avoid this, any more than you can in most other implicit differentiations. –  Brian M. Scott Jan 22 '12 at 22:12

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.

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