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

 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

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
$\cos^n x = \cos x \cdot \cos^{n1} x$
allows us to use integration by parts to get$I_n = [\sin x \cdot \cos^{n1} x]_{\pi/2}^{\pi/2} + \int_{\pi/2}^{\pi/2} (n1) \sin^2\!x \cos^{n2}\!x \,dx.$
Simplifying using a trigonometric identity yields$I_n = (n1) \int_{\pi/2}^{\pi/2} (\cos^{n2} x\cos^{n} x) \,dx,$
so$I_n = (n1)(I_{n2}I_{n}).$
Rearranging gives$I_{n} = \frac{n1}{n} I_{n2}.$
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$I_{2n} = \frac{2n1}{2n}\cdot\frac{2n3}{2n2}\cdot\frac{2n5}{2n4}\cdots\frac{1}{2}\cdot\pi$
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$I_{2n}=\pi \cdot \frac{(2n)!}{2^{2n} (n!)^2}=\binom{2n}{n}\frac{\pi}{4^n}.$