# Formal way to perform the change of variable

In solving the Van't Hoff equation

$$\frac{d\ln(K)}{dT}=\frac{\Delta H^{o}}{RT^{2}} \,\, .$$

Considering the term $\Delta H^{o}$ constant, I usually perform the separation o variables and then I perfrom an integration over the interval $[T_{1},T_{2}]$

$$\int^{T_{2}}_{T_{1}}\frac{d\ln(K)}{dT}dT=\frac{\Delta H^{o}}{R}\int^{T_{2}}_{T_{1}}T^{-2}dT \,\, .$$

My doubt lies in which is the formal way to perform a change of variables in the left hand of this equation (I'm avoiding to face that differential term as being a fraction and manipulate this such as a fraction).

My second question is how can I know if the better way to integrate this equation is defining the integration limits or just take a indefinite integral? I mean, finding the antiderivative and then findig what's the C, or defining the integration limits?

By the fundamental theorem of calculus, for any differentiable function $f$, $$\int_a^bf'(x)dx = f(b) - f(a).$$ Writing the LHS of your equation in a more suggestive form gives $$\int_{T_1}^{T_2} (\ln K)'(T) dT = \ln K(T_2) - \ln K(T_1).$$ In a more general sense, a change of variables in integration is justified by this and the chain rule $$\int_a^b f'(u(x))u'(x)dx = \int_a^b(f\circ u)'(x)dx = f(u(b)) - f(u(a)) = \int_{u(a)}^{u(b)} f'(u)du.$$ (Note the slight abuse of notation by using $u$ as the dummy variable in the last integral.) This is the rigorous justification for "cancellation" of differentials that appears when writing in Leibniz notation.
• Oh. Well that usually depends on your boundary conditions. If it's of the form $f(x_0) = y$, (which it usually is for first-order ODEs) I would suggest just doing a definite integral starting at $x_0$. Otherwise, leaving it as $C$ and figuring it out later is probably fine. Jun 27, 2017 at 21:52