Integration

$\begin{aligned} I & = \int_0^\frac \pi 4 \frac {\sin^2 5\theta}{\sin^2 \theta} - \frac {\cos^2 5\theta}{\cos^2 \theta} d \theta \\ & = \int_0^\frac \pi 4 \left(\frac {\sin 5\theta}{\sin \theta} - \frac {\cos 5\theta}{\cos \theta} \right) \left(\frac {\sin 5\theta}{\sin \theta} + \frac {\cos 5\theta}{\cos \theta} \right) d \theta \\ & = \int_0^\frac \pi 4 \frac {(\sin 5\theta \cos \theta - \cos 5 \theta \sin \theta)(\sin 5\theta \cos \theta + \cos 5 \theta \sin \theta)}{\sin^2 \theta \cos^2 \theta} d \theta \\ & = \int_0^\frac \pi 4 \frac {4\sin 4\theta \sin 6\theta }{\sin^2 2\theta} d \theta \\ & = \int_0^\frac \pi 4 \frac {8\sin 2\theta \cos 2\theta (3\sin 2\theta - 4 \sin^3 2 \theta)}{\sin^2 2\theta} d \theta \\ & = \int_0^\frac \pi 4 8 \cos 2\theta (3 - 4 \sin^2 2 \theta) \ d \theta \quad \quad \small \color{#3D99F6}{\text{Let }x = \sin 2\theta, \ dx = 2 \cos 2\theta \ d \theta} \\ & = \int_0^1 4(3-4x^2) \ dx \\ & = 12x -\frac {16x^3}3 \bigg|_0^1 = 12 - \frac {16}3 = \frac {20}3 \end{aligned}$

$\implies a+b = 20+3 = \boxed{23}$

By Euler's formula: ${\bf e^(i5\theta) = e^(i\theta)^5 \implies }$

${\bf cos(5\theta) + isin(5\theta) = (cos\theta + isin\theta)^5 = }$

${\bf cos^5\theta + 5cos^4\theta sin\theta i - 10 cos^3\theta sin^2\theta - 10cos^2\theta sin^3\theta i + 5 cos\theta sin^4\theta + sin^5\theta i \implies }$

${\bf cos(5\theta) = cos\theta (cos^4\theta - 10 cos^2\theta sin^2\theta + 5 sin^4\theta) = }$

${\bf \frac{cos\theta}{4} ((1 + cos(2\theta))^2 - 10 (1 - cos^2(2\theta)) + 5 (1 - cos(2\theta)) = }$

${\bf \frac{cos\theta}{4} (16 cos^2(2\theta) - 8 cos(2\theta) - 4) = }$

${\bf cos\theta (4 cos^2(2\theta) - 2 cos(2\theta) - 1) = }$

${\bf cos\theta (2(1 + cos(4\theta)) - 2 cos(2\theta) - 1) = }$

${\bf cos\theta (2 cos(4\theta) - 2 cos(2\theta) + 1) }$

$$

In a similiar fashion: $$

${\bf sin(5\theta) = sin\theta (2 cos(4\theta) + 2 cos(2\theta) + 1) }$

${\bf \implies }$

$\large \frac{sin^2(5\theta)}{sin^2(\theta)} - \frac{cos^2(5\theta)}{cos^2(\theta)} =$

${\bf (2 cos(4\theta) + 2 cos(2\theta) + 1)^2 - (2 cos(4\theta) - 2 cos(2\theta) + 1)^2 = }$

${\bf 8 cos(2\theta) (2 cos(4\theta) + 1) = }$

${\bf 8 cos(2\theta) ( 2 (2 cos^2(2\theta) - 1) + 1) = 8 cos(2\theta) (4 cos^2(2\theta) - 1) = }$

${\bf 8 (4 cos^3(2\theta) - cos(2\theta) = 8(4(1 - sin^2(2\theta))cos(2\theta) - cos(2\theta)) = }$

${\bf 8(3 cos(2\theta) - 4(sin(2\theta)^2 cos(2\theta)) \implies }$

$\large \int_{0}^{\frac{\pi}{4}} \frac{sin^2(5\theta)}{sin^2(\theta)} - \frac{cos^2(5\theta)}{cos^2(\theta)} d\theta =$

${\bf 12 * \int_{0}^{\frac{\pi}{4}} cos(2\theta) * 2 d\theta }$ ${\bf - 16 * \int_{0}^{\frac{\pi}{4}} (sin(2\theta))^2 2 cos(2\theta) d\theta = }$

${\bf 12 sin(2\theta) - \frac{16}{3} (sin(2\theta))^3|_{0}^{\frac{\pi}{4}} = }$

${\bf 12 - \frac{16}{3} = \frac{20}{3} = \frac{a}{b} \implies a + b = 23.}$