# 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 \(\binom{n}{k}=\frac{n!}{k!(n-k)!}.\)

## Related topics

## Warm-up Questions

- Evaluate \(\int_{-\pi}^{\pi}x^2\,dx.\)
- Find \(\int\ln x \,dx.\)
- 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 1Try to split \(\cos^n x\) into some terms, one of which you can integrate easily.
- Hint 2Have you tried integrating by parts?
- Hint 3Integrate and simplify \(I_n\) as much as you can.
- Hint 4Try to rewrite \(I_n\) as a recurrence relation.
- Hint 5... you may find a trigonometric identity helpful.
- Hint 6Which initial term can you find to help solve the recurrence for \(I_{2n}?\)
- Hint 7Try 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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