a) Find \( \int \frac{dx}{49 + x^2} \) b) Using the substitution \( u = x^2 + 8 \), or otherwise, find \( \int x \sqrt{4 + 8} \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 1 - 2006 - Paper 1

First, let’s clarify the integral to evaluate:

[ \int x \sqrt{4 + 8} , dx = \int x \sqrt{12} , dx ]

This simplifies to:

[ \int x \cdot 2\sqrt{3} , dx = 2\sqrt{3} \int x , dx = 2\sqrt{3} \cdot \frac{x^2}{2} + C = \sqrt{3} x^2 + C ]

To find this limit, we can apply L'Hôpital's rule since direct substitution results in the indeterminate form ( \frac{0}{0} ):

[ \lim_{x \to 0} \frac{\sin 5x}{3x} = \lim_{x \to 0} \frac{5\cos 5x}{3} = \frac{5\cos(0)}{3} = \frac{5}{3} ]

Using the identity for the sum of cubes:

[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) ]

Set ( a = \sin \theta ) and ( b = \cos \theta ). Thus,

[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) ]

This simplifies to:

[ \sin^2 \theta + \cos^2 \theta - \sin \theta \cos \theta = 1 - \sin \theta \cos \theta ]

Therefore, the expression becomes:

[ \frac{1 - \sin \theta \cos \theta}{\sin \theta + \cos \theta} - 1 = \frac{-\sin \theta \cos \theta - (\sin \theta + \cos \theta)}{\sin \theta + \cos \theta} = -1 ]

For the line to be tangent to the curve, both the slope and the point of intersection must match at some point ( x = a ):

1. Set the derivatives equal for tangency, meaning ( 12 = 3a^2 ) leading to ( a^2 = 4 ), so ( a = 2 ) or ( a = -2 ).
2. Substitute these values back into the function:
   • For ( a = 2 ): ( y = 12 \cdot 2 + b = 24 + b ) and ( y = 2^3 = 8 ) implies ( b = 8 - 24 = -16 ).
   • For ( a = -2 ): ( y = 12 \cdot (-2) + b = -24 + b ) and ( y = (-2)^3 = -8 ) implies ( b = -8 + 24 = 16 ).

Thus, the values of ( b ) are ( -16 ) and ( 16 ).