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# Practice Paper 3 Question 11

Let $$I_n=\int_{-\pi/2}^{\pi/2} \cos^n\!x \,dx$$ for any non-negative integer $$n.$$ $$\;(i)$$ Find a recursive expression for $$I_n.$$ $$(ii)$$ Find the simplest non-recursive expression for $$I_{2n}$$ that contains $$\binom{2n}{n},$$ where $$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$$

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## Warm-up Questions

1. Evaluate $$\int_{-\pi}^{\pi}x^2\,dx.$$
2. Find $$\int\ln x \,dx.$$
3. A geometric sequence starts with $$\frac{1}{5}, \frac{2}{15}, \frac{4}{45} \ldots.$$ Find a recursive formula for the sequence and solve it to get the closed form.

## Hints

• Hint 1
Try to split $$\cos^n x$$ into some terms, one of which you can integrate easily.
• Hint 2
Have you tried integrating by parts?
• Hint 3
Integrate and simplify $$I_n$$ as much as you can.
• Hint 4
Try to rewrite $$I_n$$ as a recurrence relation.
• Hint 5
... you may find a trigonometric identity helpful.
• Hint 6
Which initial term can you find to help solve the recurrence for $$I_{2n}?$$
• Hint 7
Try writing out and unwinding $$I_{2n}.$$
• Hint 8
... remembering that your expression should contain $$\frac{(2n)!}{(n!)^2}.$$

## Solution

Writing $$\cos^n x = \cos x \cdot \cos^{n-1} x$$ allows us to use integration by parts to get $$I_n = [\sin x \cdot \cos^{n-1} x]_{-\pi/2}^{\pi/2} + \int_{-\pi/2}^{\pi/2} (n-1) \sin^2\!x \cos^{n-2}\!x \,dx.$$ Simplifying using a trigonometric identity yields $$I_n = (n-1) \int_{-\pi/2}^{\pi/2} (\cos^{n-2} x-\cos^{n} x) \,dx,$$ so $$I_n = (n-1)(I_{n-2}-I_{n}).$$ Rearranging gives $$I_{n} = \frac{n-1}{n} I_{n-2}.$$

To solve this recurrence relation for $$I_{2n},$$ we first find $$I_0=\int_{-\pi/2}^{\pi/2}dx=\pi.$$ Unwinding the recursion we get $$I_{2n} = \frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdots\frac{1}{2}\cdot\pi$$ which has the product of all odd terms in the numerator and the product of all even terms in the denominator. To get $$(2n)!$$ in the numerator, we need to fill in the even terms, so we multiply both the numerator and denominator with the denominator, which yields $$I_{2n} = \pi \cdot \frac{(2n)!}{2^2 4^2 6^2 \cdots (2n)^2}.$$ Factorising $$2^2$$ from each term in the denominator gives $$I_{2n}=\pi \cdot \frac{(2n)!}{2^{2n} (n!)^2}=\binom{2n}{n}\frac{\pi}{4^n}.$$

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