Draw the truth table for each of the propositions p q and p q and show that they are identical. Also, given that ^-1(x) = alpha and ^-1(x) = beta, show that (alpha + beta) = 1.

Here are the solutions to the problems:

3. (i) Draw the truth table for each of the propositions $p \implies q$ and $\sim p \lor q$ and show that they are identical.

Step 1: Create a truth table for $p$ and $q$.

Step 2: Determine the truth values for $p \implies q$.

Step 3: Determine the truth values for $\sim p$.

Step 4: Determine the truth values for $\sim p \lor q$.

Step 5: Compare the columns for $p \implies q$ and $\sim p \lor q$.

$$\begin{array}{|c|c|c|c|c|} \hline p & q & p \implies q & \sim p & \sim p \lor q \\ \hline T & T & T & F & T \\ T & F & F & F & F \\ F & T & T & T & T \\ F & F & T & T & T \\ \hline \end{array}$$

The columns for $p \implies q$ and $\sim p \lor q$ are identical, which shows that the two propositions are equivalent.

3. (ii) Given that $\sin^{-1}(x) = \alpha$ and $\cos^{-1}(x) = \beta$, show that $\sin(\alpha + \beta) = 1$.

Step 1: Use the property of inverse trigonometric functions.
We know that for any $x \in [-1, 1]$, the sum of $\sin^{-1}(x)$ and $\cos^{-1}(x)$ is $\frac{\pi}{2}$.

Step 2: Substitute the given values into the identity.
Given $\sin^{-1}(x) = \alpha$ and $\cos^{-1}(x) = \beta$, we have:

$$\alpha + \beta = \frac{\pi}{2}$$

Step 3: Calculate $\sin(\alpha + \beta)$.

$$\sin(\alpha + \beta) = \sin\left(\frac{\pi}{2}\right)$$

Step 4: Evaluate the sine function.

$$\sin\left(\frac{\pi}{2}\right) = 1$$

Thus, $\sin(\alpha + \beta) = 1$.

4. (i) The function $f$ is given by $f(x) = \frac{x+1}{(x-1)(x^2+1)}$. Express $f(x)$ in partial fractions. Hence find $\int f(x)\,dx$.