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Let A_n = \int_0^{\frac{\pi}{2}} \cos^{2n} x \, dx and B_n = \int_0^{\frac{\pi}{2}} x^2 \cos^{2n} x \, dx , where n is an integer, n \geq 0 - HSC - SSCE Mathematics Extension 2 - Question 8 - 2010 - Paper 1
Question 8
Let A_n = \int_0^{\frac{\pi}{2}} \cos^{2n} x \, dx and B_n = \int_0^{\frac{\pi}{2}} x^2 \cos^{2n} x \, dx , where n is an integer, n \geq 0. (Not... show full transcript
Worked Solution & Example Answer:Let A_n = \int_0^{\frac{\pi}{2}} \cos^{2n} x \, dx and B_n = \int_0^{\frac{\pi}{2}} x^2 \cos^{2n} x \, dx , where n is an integer, n \geq 0 - HSC - SSCE Mathematics Extension 2 - Question 8 - 2010 - Paper 1
Step 1
Show that n A_n = \frac{2n - 1}{2} A_{n - 1} for n \geq 1.
Answer
Using the reduction formula for integrals of cosine, we can evaluate (A_n) as follows:
Let's start with the integral:
Using integration by parts, let:
- u = \cos^{2n-1} x \quad \Rightarrow \quad du = -(2n-1) \cos^{2n-2} x \sin x , dx
- dv = \cos x , dx \quad \Rightarrow \quad v = \sin x
Now applying the integration by parts formula ( \int u , dv = uv - \int v , du ):
The first term evaluates to zero, resulting in:
Thus, isolating the term gives the desired result:
Step 2
Using integration by parts on A_n, or otherwise, show that A_n = 2n \int_0^{\frac{\pi}{2}} x \sin x \cos^{2n - 1} x \, dx for n \geq 1.
Answer
Using the result from part (a), we can write:
This comes from applying integration by parts to the expression derived previously, substituting and rearranging appropriately.
Step 3
Use integration by parts on the integral in part (b) to show that \frac{A_n}{n^2} = \frac{(2n - 1)}{2} B_{n - 1} - 2 B_n for n \geq 1.
Answer
For part (c), we take:
Applying integration by parts again:
- Let u = x and dv = \sin x \cos^{2n-1} x , dx.
We find:
Using previous results, we substitute back to get:
Step 4
Use parts (a) and (c) to show that \frac{1}{n^2} = \frac{B_{n - 1}}{A_{n - 1}} - \frac{B_n}{A_n} for n \geq 1.
Answer
By using the equations derived in parts (a) and (c), we have:
-
From part (a):
(n A_n = \frac{2n - 1}{2} A_{n - 1}) -
From part (c):
(\frac{A_n}{n^2} = \frac{(2n - 1)}{2} B_{n - 1} - 2 B_n)
Substituting these into the equations, it simplifies to show that:
Step 5
Show that \sum_{k=1}^{n} \frac{1}{k^2} = \frac{\pi^2}{6} - \frac{B_n}{A_n}.
Answer
Taking our results from part (d), we relate the series to known results about the sums of reciprocals of squares:
By observing the series sum involves the value we derived, we write:
This identifies the relationship needed.
Step 6
Use the fact that \frac{n}{A_n} \geq \frac{2}{\pi} for 0 \leq x \leq \frac{\pi}{2} to show that B_n \geq \int_0^{\frac{\pi}{2}} \left( 1 - \frac{4x^2}{\pi^2} \right)^n dx.
Answer
Using the known bound on (A_n), we can substitute into the expression for (B_n).
By analyzing the inequalities, derive:
Thus, showing that our statement holds.