$\int_0^{\pi /4} \tan^2 x \, dx = $

Let $f$ be a real-valued continuous function defined on the positive real axis such that $g(x) = \int_0^x t f(t) dt$. If $g(x^3) = x^6 + x^7$,then the value of $\sum_{r=1}^{15} f(r^3)$ is:

If $f(x) = \begin{cases} \frac{6 x^2 + 1}{4 x^3 + 2 x + 3}, & 0 < x < 1 \\ x^2 + 1, & 1 \le x \le 2 \end{cases}$ then $\int_0^2 f(x) dx =$

$\int_0^\pi \frac{dx}{1 + \sin x} = $

The approximate value of $\int_{1}^{9} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $