Given two problems:

Problem A: Tau = Integral(0 to v)
v/(q^2 - v^2) dv
q = constant

Problem B: Tau = Integral(V TO 0)
v/(q^2 - v^2) dv
q = constant

Why is it that, SUBSTITUTION RULE is used in problem A, and QUOTIENT RULE is used in problem B? The two problems are somewhat similar. The only different is their integrals. Problem A, the integration is from 0 to V and problem B, the integration is from V to 0. The answers are shown below. Please help!!

Answer (problem A):
Tau = ln(q^2/(q^2-v^2))

Answer (problem B):
Tau = ln((q^2-v^2)/q^2)

One integral is the negative of the other, since the integrand is the same and only the direction of integration changes. If you think about the answers, and recognize that log a/b = - log b/a, you will see that all that is involved is a sign change there also. Either integration could have been solved by either method.

In problem A, the substitution rule is used because the limits of integration are changing. The integral is taken from 0 to v, which means that the variable of integration changes from the lower limit to the upper limit. To handle this change in limits, a substitution is made. Let's see how it is done.

First, we let u = q^2 - v^2. Taking the derivative of u with respect to v, we have du/dv = -2v. Solving for dv, we have dv = -du/(2v).

Now, let's find the limits of integration in terms of u. When v = 0, u = q^2 - 0^2 = q^2. When v = v, u = q^2 - v^2.

Next, substitute the expression for v and dv into the integral:

Tau = Integral(0 to v) v/(q^2 - v^2) dv

Since v/(q^2 - v^2) is v/u and dv is -du/(2v), the integral becomes:

Tau = Integral(q^2 to u) -du/2u

Now, we can solve the integral using the basic integral rule for u:

Tau = (-1/2) * ln|u| + C

Substituting back u = q^2 - v^2:

Tau = (-1/2) * ln|q^2 - v^2| + C

Finally, we can simplify the absolute value notation by recognizing that q^2 - v^2 is always positive (since q is constant). Thus, we have:

Tau = -1/2 * ln(q^2 - v^2) + C

Therefore, the answer for problem A is Tau = ln(q^2/(q^2 - v^2)).

Now, let's move on to problem B where the integration is taken from V to 0.

The quotient rule is used in problem B because we are directly integrating the equation without any change in limits. The variable of integration is v itself.

Again, let's write down the integral:

Tau = Integral(V to 0) v/(q^2 - v^2) dv

In this case, we don't need to make any substitutions. We can use the quotient rule directly.

Let's decompose the integrand into partial fractions:

v/(q^2 - v^2) = A/(q+v) + B/(q-v)

Cross multiplying and combining like terms, we get:

v = A(q-v) + B(q+v)

Expanding and collecting terms, we have:

v = (A+B)q + (-A+B)v

Comparing the coefficients of q and v on both sides, we get:

A + B = 0 --> B = -A
A - B = 1 --> 2A = 1 --> A = 1/2

Substituting A = 1/2 back into B = -A, we find B = -1/2.

Now, rewrite the integrand in terms of A and B:

v/(q^2 - v^2) = 1/2(q+v)/(q+v) - 1/2(q-v)/(q-v)

Simplifying further, we have:

v/(q^2 - v^2) = 1/2 + (v-q)/(2(q-v))

Now, let's integrate this expression:

Tau = Integral(V to 0) v/(q^2 - v^2) dv
= 1/2 * Integral(V to 0) dv + 1/2 * Integral(V to 0) (v-q)/(q-v) dv

The first integral is straightforward and gives us:

Tau = 1/2 * [v] (evaluated from V to 0)
= 1/2 * (0 - V) = -1/2 * V

For the second integral, we can use the logarithm properties:

Integral(V to 0) (v-q)/(q-v) dv = -Integral(0 to V) (v-q)/(v-q) dv
= -Integral(0 to V) dv
= -[v] (evaluated from 0 to V)
= -(V - 0) = -V

Substituting both results back into Tau, we have:

Tau = -1/2 * V - (-V) = -1/2 * V + V = 1/2 * V

Therefore, the answer for problem B is Tau = ln((q^2 - v^2)/q^2).

I hope this explanation helps! Let me know if you have any further questions.