Answer
Using the Integration by Parts Formula in Section 8.1, we verify Eq. (8):
${C_n} = \left( {\frac{{n - 1}}{n}} \right){C_{n - 2}}$.
Work Step by Step
Recall the definition: ${C_n} = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^n}\theta {\rm{d}}\theta $.
First, write
$\smallint {\cos ^n}\theta {\rm{d}}\theta = \smallint \cos \theta {\cos ^{n - 1}}\theta {\rm{d}}\theta $
Since $\cos \theta {\rm{d}}\theta = {\rm{d}}\left( {\sin \theta } \right)$. So,
$\smallint {\cos ^n}\theta {\rm{d}}\theta = \smallint {\cos ^{n - 1}}\theta {\rm{d}}\left( {\sin \theta } \right)$
Write $u = {\cos ^{n - 1}}\theta $ and $v = \sin \theta $. Using the Integration by Parts Formula in Section 8.1, we obtain
$\smallint {\cos ^n}\theta {\rm{d}}\theta = \left( {{{\cos }^{n - 1}}\theta } \right)\left( {\sin \theta } \right) - \smallint sin\theta {\rm{d}}\left( {{{\cos }^{n - 1}}\theta } \right)$
Since ${\rm{d}}\left( {{{\cos }^{n - 1}}\theta } \right) = - \left( {n - 1} \right)\left( {\sin \theta } \right)\left( {{{\cos }^{n - 2}}\theta } \right){\rm{d}}\theta $. So,
$\smallint {\cos ^n}\theta {\rm{d}}\theta = \left( {{{\cos }^{n - 1}}\theta } \right)\left( {\sin \theta } \right) + \left( {n - 1} \right)\smallint {\sin ^2}\theta {\cos ^{n - 2}}\theta {\rm{d}}\theta $
Using ${\sin ^2}\theta = 1 - {\cos ^2}\theta $, we get
$\smallint {\cos ^n}\theta {\rm{d}}\theta = \left( {{{\cos }^{n - 1}}\theta } \right)\left( {\sin \theta } \right) + \left( {n - 1} \right)\left( {\smallint {{\cos }^{n - 2}}\theta {\rm{d}}\theta - \smallint {{\cos }^n}\theta {\rm{d}}\theta } \right)$
There is the term $\smallint {\cos ^n}\theta {\rm{d}}\theta $ on both sides, so
$\left( {1 + n - 1} \right)\smallint {\cos ^n}\theta {\rm{d}}\theta = \left( {{{\cos }^{n - 1}}\theta } \right)\left( {\sin \theta } \right) + \left( {n - 1} \right)\smallint {\cos ^{n - 2}}\theta {\rm{d}}\theta $
$n\smallint {\cos ^n}\theta {\rm{d}}\theta = \left( {{{\cos }^{n - 1}}\theta } \right)\left( {\sin \theta } \right) + \left( {n - 1} \right)\smallint {\cos ^{n - 2}}\theta {\rm{d}}\theta $
Write the definite integral:
$n\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^n}\theta {\rm{d}}\theta = \left( {{{\cos }^{n - 1}}\theta } \right)\left( {\sin \theta } \right)|_{ - \pi /2}^{\pi /2} + \left( {n - 1} \right)\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^{n - 2}}\theta {\rm{d}}\theta $
The first term on the right-hand side is zero, so
$n\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^n}\theta {\rm{d}}\theta = \left( {n - 1} \right)\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^{n - 2}}\theta {\rm{d}}\theta $
$\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^n}\theta {\rm{d}}\theta = \left( {\frac{{n - 1}}{n}} \right)\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^{n - 2}}\theta {\rm{d}}\theta $
By definition:
${C_n} = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^n}\theta {\rm{d}}\theta $
and
${C_{n - 2}} = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^{n - 2}}\theta {\rm{d}}\theta $
Hence, Eq. (8):
${C_n} = \left( {\frac{{n - 1}}{n}} \right){C_{n - 2}}$.
