The value of the integral $I = \int_{0}^{1} x(1 - x)^n dx$ is

If ${I_n} = \int_{0}^{\pi /4} {\tan^n x} \,dx$,then $\lim_{n \to \infty} n[{I_n} + {I_{n - 2}}]$ equals

Assertion $(A): \int_{-a}^a f(x) dx = \int_0^a (f(x) + f(-x)) dx$
Reason $(R): \int_a^b f(x) dx = \int_{g(a)}^{g(b)} f(g(u)) g'(u) du$
The correct option among the following is:

$\int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x=$

$\int_{ - \pi }^{\pi } {\frac{{2x(1 + \sin x)}}{{1 + {{\cos }^2}x}}dx} $ is

If $\int_{-1}^4 f(x) dx = 4$ and $\int_2^4 (3 - f(x)) dx = 7$,then $\int_{-1}^2 f(x) dx = $