Let $f(x) = \begin{cases} \frac{1}{3}, & x \le \pi/2 \\ \frac{b(1-\sin x)}{(\pi-2x)^2}, & x > \pi/2 \end{cases}$. If $f$ is continuous at $x = \pi/2$, then the value of $\int_0^{3b-6} |x^2+2x-3| dx$ is:

If $\int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \,d x=\frac{k}{6}$, then the value of $k$ is

Let $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$,for all real $x$. Then,$\lim_{x \rightarrow 0} \frac{g(x)}{x}$ is equal to

Let $f(x) = 2 + |x| - |x - 1| + |x + 1|$,$x \in R$. Consider:
$(S1): f^{\prime}\left(-\frac{3}{2}\right) + f^{\prime}\left(-\frac{1}{2}\right) + f^{\prime}\left(\frac{1}{2}\right) + f^{\prime}\left(\frac{3}{2}\right) = 4$
$(S2): \int_{-2}^{2} f(x) dx = 12$
Then,

The value of $\int_0^\pi |\sin^3 x| dx$ is

$\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}$ is equal to