$\int_{0}^{1} x(1-x)^{5} dx =$

Show that $\int_{0}^{a} f(x) g(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$,if $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$.

If $I_n = \int_0^{\frac{\pi}{4}} \tan^n \theta \, d\theta$,then $I_{12} + I_{10} =$

If for non-zero $x,$ $af(x) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5,$ where $a \ne b,$ then $\int_1^2 {f(x)\,dx = } $

Let the function $F$ be defined as $F(x) = \int_{1}^{x} \frac{e^{t}}{t} dt, x > 0$. Then the value of the integral $\int_{1}^{x} \frac{e^{t}}{t+a} dt$,where $a > 0$,is

$\frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx =$