Evaluate \( \int_0^{\frac{\pi}{2}} \cos x \sin^2 x \, dx - HSC - SSCE Mathematics Extension 1 - Question 4 - 2005 - Paper 1

Using the substitution ( t = \tan \frac{\theta}{2} ), we know that:

$\csc \theta = \frac{1 + t^2}{2t} \quad \text{and} \quad \cot \theta = \frac{1 - t^2}{2t} .$

Adding these gives:

$\csc \theta + \cot \theta = \frac{1 + t^2 + (1 - t^2)}{2t} = \frac{2}{2t} = \frac{1}{t} = \cot \frac{\theta}{2} .$

Let the equations of the normals be:

1. ( x + py - 2ap + q^3 = 0 )
2. ( x + qy - 2aq + p^3 = 0 )

By solving these two equations simultaneously, we can find the coordinates of the intersection point ( R ).

From the first equation, we can express ( x ) in terms of ( y ): [ x = -py + 2ap - q^3 . ] Substituting this in the second equation results in: [ -py + 2ap - q^3 + qy - 2aq + p^3 = 0 . ]

Rearranging gives us a system of equations that leads to the coordinates for ( R ). After simplification, we find that [ R = \left( -apq[p + q], a \left[ b^2 + pq + q^2 + 2 \right] \right) . ]

The equation of the chord can be derived using the two-point form of a line, but the exact steps are not required here.

Substituting ( (0, a) ) into the equation of the chord gives: [ a = \frac{1}{2}(p + q)(0) - apq , ] which simplifies to: [ a + apq = 0 . ] If we rearrange, we have: [ apq = -a . ] Since ( a \neq 0 ), we divide to get: [ pq = -1 . ]

Determining the locus involves eliminating parameters from the equation of the chord and the coordinates of the point ( R ). By substituting ( pq = -1 ) into the expression for the coordinates of ( R ), we can derive an equation involving only the coordinates, leading to the locus of ( R ) in terms of ( a ) and other constants.

To prove this statement using induction, we first verify the base case for ( n = 2 ):

$4^2 - 1 - 7(2) = 16 - 1 - 14 = 1 > 0 .$

Now we assume it holds for some integer ( k ), i.e., ( 4^k - 1 - 7k > 0 . )

We need to show it holds for ( k + 1 ):

$4^{k + 1} - 1 - 7(k + 1) = 4(4^k) - 1 - 7k - 7 = 4(4^k - 1 - 7k) + 3 .$

By the induction hypothesis, since ( 4^k - 1 - 7k > 0 ), thus ( 4(4^k - 1 - 7k) + 3 > 3 > 0 . $$ This completes the induction step.