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

Try to split $\cos^n x$ into some terms, one of which you can integrate easily.

Have you tried integrating by parts?

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.

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 $\frac{(2n)!}{(n!)^2}.$

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 $I_{2n}=\pi \cdot \frac{(2n)!}{2^{2n} (n!)^2}=\binom{2n}{n}\frac{\pi}{4^n}.$