Evaluate the definite integral: $\int_{0}^{3} \sqrt{9 - x^2} dx$.

Let $f(x) = \{x\}$ denote the fractional part of a real number $x$. Then, the value of $\int_{0}^{\sqrt{3}} f(x^2) dx$ is

If $[x]$ denotes the greatest integer less than or equal to $x$,then the value of $\int_{1}^{5} [|x - 3|] \, dx$ is

If $5 f(x)+3 f\left(\frac{1}{x}\right)=2-\frac{1}{x}, x \neq 0$,then $\int_1^2 f\left(\frac{1}{x}\right) d x=$

$\int_3^8 \frac{2 - 3x}{x\sqrt{1 + x}} \, dx$ is equal to

If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$,then the value of the integral $\int_{0}^{\theta} \sin x \,dx$ is equal to: