Which of these integrals converge ?
I am confused about how to check for the convergence when the functions are more complex inside the integral.
My attempt:
in option C : integrating gives -2 and hence should converge i guess. in option D: its sin inverse (x) and the upper limit 2 actually cant be applied right ? option a and option b i am pretty much clueless on how to even start with?!
a) converge $\int=2$.
b) converge because $-1\le \sin(x) \le 1$ so $\int \le \int_1^\infty \frac1{x^3}dx$ which converge.
c) not converge : 0 is a singularity of $1/x^2$ and is in the integration interval. and the two integrals $\int_{-1}^0$ , $\int_0^1$ are not converging.
d) 1 which is a singularity of $f(x)=\frac1{\sqrt{|1-x^2|}}$. But, the two integrals \int_0^1 f(x) dx and $\int_1^2 f(x) dx are converging. So the integral converge.
the primitive of $\frac1{\sqrt{1-x^2}}$ is $\arcsin(x)$ ,and of $\frac1{\sqrt{x^2-1}}$ is $\operatorname{arcsinh}(x)$.