11. Find \( \int \sin^3 x \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

1. Identify the slopes of the lines. The slope of ( y = 2x + 5 ) is ( m_1 = 2 ) and the slope of ( y = 4 - 3x ) is ( m_2 = -3 ).

2. Use the formula for the angle between two lines: [ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} = \frac{2 - (-3)}{1 + (2)(-3)} = \frac{5}{-5} = -1. ]

3. Thus, ( \theta = \tan^{-1}(-1), ) indicating an angle of ( 135^{\circ} ).

4. The size of the acute angle is then ( 180^{\circ} - 135^{\circ} = 45^{\circ} ) or ( \frac{\pi}{4} ) radians.

1. Start by rearranging the inequality: [ \frac{4}{x+3} - 1 \geq 0 \Rightarrow \frac{4 - (x + 3)}{x + 3} \geq 0 \Rightarrow \frac{1 - x}{x + 3} \geq 0. ]

2. Identify the critical points where the numerator and denominator are zero:

   • Numerator: ( 1 - x = 0 ) gives ( x = 1 ).
   • Denominator: ( x + 3 = 0 ) gives ( x = -3 ).

3. Create a number line and test intervals defined by these points to find the solution set:

   • Test interval ( (-\infty, -3) ) gives a positive value.
   • Test interval ( (-3, 1) ) gives a negative value.
   • Test interval ( (1, \infty) ) gives a negative value.

4. Therefore, the solution is ( x < -3 ) or ( x = 1 ).

1. Use the transformation for expressions of this form: [ A \cos(x + \alpha) = A \cos x \cos \alpha - A \sin x \sin \alpha. ]

2. Compare coefficients: Set ( A \cos \alpha = 5 ) and ( A \sin \alpha = 12 ).

3. To find ( A ), calculate: [ A^2 = (5)^2 + (12)^2 = 25 + 144 = 169 \Rightarrow A = 13. ]

4. Find ( \alpha ) using: [ \tan \alpha = \frac{12}{5}, \text{ thus } \alpha = \tan^{-1}\left(\frac{12}{5}\right). ]

The expression is thus ( 13 \cos\left(x + \tan^{-1}\left(\frac{12}{5}\right)\right) ).

1. Change the variable using the substitution ( u = 2x - 1 ) which implies ( x = \frac{u + 1}{2} ) and ( dx = \frac{1}{2} du. )

2. Rewrite the limits: When ( x = 1, u = 1 ), and when ( x = 2, u = 3 ).

3. Substitute into the integral: [ \int_{1}^{3} \frac{\frac{u + 1}{2}}{u^3} \cdot \frac{1}{2} du = \frac{1}{4} \int_{1}^{3} \frac{u + 1}{u^3} , du. ]

4. Split the integrals: [ = \frac{1}{4} \left(\int_{1}^{3} u^{-2} , du + \int_{1}^{3} u^{-3} , du\right) ] [ = \frac{1}{4} \left(-u^{-1}\bigg|_1^3 + \frac{1}{2}u^{-2}\bigg|_1^3\right). ]

5. Evaluate to find the result.

1. Since ( A(x) = x - 3 ) is a factor, substituting ( x = 3 ) into ( P(x) ) must yield zero: [ P(3) = 3^3 - k(3^2) + 5(3) + 12 = 0. ] [ 27 - 9k + 15 + 12 = 0 \Rightarrow 54 - 9k = 0 \Rightarrow k = 6. ]

1. Substitute ( k = 6 ) into ( P(x) ): [ P(x) = x^3 - 6x^2 + 5x + 12. ]

2. Use synthetic division to find other roots or factor: [ P(x) = (x - 3)(x^2 - 3x - 4). ]

3. Factor further: [ = (x - 3)(x - 4)(x + 1). ]

The zeros of ( P(x) ) are ( x = 3, 4, -1. ]