Definitely Integrating

As was hinted, let $u = sin(\theta)$ . Then, $\frac{du}{d\theta} = cos(\theta)$ and $d(\theta) = \frac{du}{cos(\theta)}$ .

Now, $\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 4 } }{ cos\theta (u -1)^{ 3 }\quad \frac{du}{cos(\theta)} }$

= $\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 4 } }{ (u -1)^{ 3 }\quad du }$

Applying the Power Rule $\int x^{a}dx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

Now $[\frac{\left(u-1\right)^4}{4}] = [\frac{\left(sin(\theta)-1\right)^4}{4}]$

Definite Integration:

$\frac{\left(sin(\frac { \pi }{ 4} )-1\right)^4}{4}- \frac{\left(sin(\frac { \pi }{ 6 } )-1\right)^4}{4}$

$= \frac{\left((\frac { \sqrt2}{2 }) -1\right)^4}{4}- \frac{\left((0.5)-1\right)^4}{4}$

$\boxed{\approx -0.0138}$