$\int \frac{\sec x}{3(\sec x+\tan x)+2} d x=$

Let $\int \frac{2-\tan x}{3+\tan x} dx = \frac{1}{2}(\alpha x + \log_e |\beta \sin x + \gamma \cos x|) + C$,where $C$ is the constant of integration. Then $\alpha + \frac{\gamma}{\beta}$ is equal to:

If $\int(\sin x )^{\frac{-11}{2}}(\cos x )^{\frac{-5}{2}} dx = -\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}}-\frac{p_2}{q_2}(\cot x)^{\frac{5}{2}}-\frac{p_3}{q_3}(\cot x)^{\frac{1}{2}}+\frac{p_4}{q_4}(\cot x)^{\frac{-3}{2}}+C,$ where $p_i$ and $q_i$ are positive integers with $\operatorname{gcd}(p_i, q_i)=1$ for $i =1,2,3,4$ and $C$ is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to . . . . . . .

The function $f(x)$,that satisfies the condition $f(x)=x+\int_{0}^{\pi / 2} \sin x \cdot \cos y f(y) dy$,is :

$\int \frac{d x}{\left(2 a x+x^2\right)^{\frac{3}{2}}} = $

For $n \geq 2$,let $I_n = \int_0^{\pi / 4} \tan^n x \, dx$ and $F_n = I_n + I_{n-2}$. Then,$F_n - F_{n+1} =$