$I_n=\int_{-\pi/2}^{\pi/2} \cos^n\!x \,dx$ 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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- 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.
- Evaluate
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Try to split
$\cos^n x$into some terms, one of which you can integrate easily. -
Have you tried integrating by parts?
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Integrate and simplify
$I_n$as much as you can. -
Try to rewrite
$I_n$as a recurrence relation. -
… you may find a trigonometric identity helpful.
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Which initial term can you find to help solve the recurrence for
$I_{2n}?$ -
Try writing out and unwinding
$I_{2n}.$ -
… remembering that your expression should contain
$\frac{(2n)!}{(n!)^2}.$
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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$I_{2n}=\pi \cdot \frac{(2n)!}{2^{2n} (n!)^2}=\binom{2n}{n}\frac{\pi}{4^n}.$