If $I_1 = \int_0^{\pi / 4} \sin^2 x \, dx$ and $I_2 = \int_0^{\pi / 4} \cos^2 x \, dx$, then,

Let $f:[-1, 2] \rightarrow [0, \infty)$ be a continuous function such that $f(x) = f(1-x)$ for all $x \in [-1, 2]$. Let $R_1 = \int_{-1}^2 x f(x) dx$,and $R_2$ be the area of the region bounded by $y = f(x)$,$x = -1$,$x = 2$,and the $x$-axis. Then

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4 x} dx = $ . . . . . . .

$\int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(588 - 84x + 3x^2)} dx =$

By using the properties of definite integrals,evaluate the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x \, dx$.

If $g(x) = \int_0^x \cos^4 t \,dt$, then $g(x+\pi)$ equals