QUESTION 4: INTEGRATION 1 Simplify the following: 2 Integrate the following: 3 Evaluate the following:

Let $u = x^2$ and $v = \ln 3x$.

The answer is in positive exponents and does not contain surds.

3.4 Determine the co-ordinates of the points of inflection for $y = \sin x$; $0^\circ \le x \le 270^\circ$ with the AID of DIFFERENTIATION.

For the given interval $0^\circ \le x \le 270^\circ$, the values of $x$ for which $\sin x = 0$ are $x = 0^\circ$ and $x = 180^\circ$.

• At $x = 0^\circ$: For $x > 0^\circ$ (e.g., $x=1^\circ$), $\sin x > 0$, so $\frac{d^2y}{dx^2} = -\sin x < 0$. The graph is concave down. Since $0^\circ$ is an endpoint, concavity change cannot be fully assessed.
• At $x = 180^\circ$: For $x < 180^\circ$ (e.g., $x=170^\circ$), $\sin x > 0$, so $\frac{d^2y}{dx^2} = -\sin x < 0$. (Concave down) For $x > 180^\circ$ (e.g., $x=190^\circ$), $\sin x < 0$, so $\frac{d^2y}{dx^2} = -\sin x > 0$. (Concave up) Since the concavity changes at $x = 180^\circ$, this is a point of inflection.

Substitute $x = 180^\circ$ into the original equation $y = \sin x$:

The point of inflection is $\boxed{(180^\circ, 0)}$.

Question 4: INTEGRATION

4.1 Simplify the following: $\int (\sqrt{1 + \cot^2 x} \cdot \cot x)\,dx$

Assuming $x$ is in a domain where $\csc x > 0$, we have $|\csc x| = \csc x$.

Recall that the derivative of $\csc x$ is $-\csc x \cot x$.
Therefore, the integral of $\csc x \cot x$ is $-\csc x$.

4.2 Integrate the following: $\int \left(e^{-3x} + \frac{\tan x + \sin 2x}{\cos x} + \sqrt{x} + 2t^3 + \frac{3}{x} + \theta x\right)\,dx$

• For $e^{-3x}$:

• For $\frac{\tan x + \sin 2x}{\cos x}$: First, simplify the expression:

Now integrate:

• For $\sqrt{x}$:

• For $2t^3$: (assuming $t$ is a constant with respect to $x$)

• For $\frac{3}{x}$:

• For $\theta x$: (assuming $\theta$ is a constant with respect to $x$)

4.3 Evaluate the following:

4.3.1 $\int_1^2 (3t^2 + 2t)\,dt$

4.3.2 $\int_0^{\pi/2} \sqrt{\cos^2 \theta}\,d\theta$

For the interval $0 \le \theta \le \frac{\pi}{2}$, $\cos \theta$ is non-negative.
So, $|\cos \theta| = \cos \theta$.
The integral becomes: