For $m,n>0$, let $\alpha(m,n)=\displaystyle\int_0^2 t^m(1+3t)^n\,dt$. If $11\alpha(10,6)+18\alpha(11,5)=p(14)^6$, then $p$ is equal to
Let $\beta(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,dx$, $m,n>0$. If $\int_0^1(1-x^{10})^{20}\,dx=a\cdot\beta(b,c)$, then $100(a+b+c)$ equals:
Question 25: Let \(I = \int \tan^6 x\, dx - \int \tan^4 x(\sec^2 x - 1)\, dx\). If \(I = \frac{\tan^5 x}{5} - \frac{\tan^3 x}{3} + A\tan x - x + D\) where \(A, B, C\) are constants with \(A = -\frac{1}{3}, B = 1, C = -1\), then \(A + B + C\) equals: