Problem $2\mathrm{.}2$

Show that for odd powers of cosine,

$I_{2n+1}={\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}co{s}^{2n+1}xdx=\frac{2*4*6\text{c}\text{d}\text{o}\text{t}\text{s}(2n)}{3*5*7\text{c}\text{d}\text{o}\text{t}\text{s}(2n+1)}$

Hint: Using induction, we can assume that the previous case is true:

$I_{2n-1}={\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}co{s}^{2n-1}xdx=\frac{2*4*6\text{c}\text{d}\text{o}\text{t}\text{s}(2n-2)}{3*5*7\text{c}\text{d}\text{o}\text{t}\text{s}(2n-1)}$

Then use the recursion formula to find $I_{2n+1}.$Problem $2\mathrm{.}2$

Show that for odd powers of cosine,

$I_{2n+1}={\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}co{s}^{2n+1}xdx=\frac{2*4*6\text{c}\text{d}\text{o}\text{t}\text{s}(2n)}{3*5*7\text{c}\text{d}\text{o}\text{t}\text{s}(2n+1)}$

Hint: Using induction, we can assume that the previous case is true:

$I_{2n-1}={\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}co{s}^{2n-1}xdx=\frac{2*4*6\text{c}\text{d}\text{o}\text{t}\text{s}(2n-2)}{3*5*7\text{c}\text{d}\text{o}\text{t}\text{s}(2n-1)}$

Then use the recursion formula to find $I_{2n+1}.$