Some theory :-
$p-$test for discontinuity :-. For $a>0$ ,
- If $p<1$, then $\large \int_{0}^{a}\frac{1}{x^{p}}dx$ converges .
- If $p \geq 1$, then $\large \int_{0}^{a}\frac{1}{x^{p}}dx$ diverges .
$p-$test for Infinite limits :-. For $a>0$ ;
- If $p>1$, then $\large \int_{a}^{\infty}\frac{1}{x^{p}}dx$ converges .
- If $p \leq 1$, then $\large \int_{a}^{\infty}\frac{1}{x^{p}}dx$ diverges .
Option 1:-
$\large \int_{0}^{1}\frac{\sin x }{1-\cos x} dx$
let , $\large z=1-\cos x$
$\large dz=\sin x dx$
Now our integral looks like ,
$\large \int_{0}^{0.459}\frac{1}{z}dz$
Now if we compare it with $p-$test for discontinuity 2nd part , here p=1 a>0 which match the condition .
So this improper integral diverges .
Option 3:-
$\large \int_{0}^{\infty}\frac{x}{1+x^{2}}dx$
Let ,$z=1+x^2$
$dz=2x dx$
Now our integral looks like ,
$\large \frac{1}{2}\int_{1}^{\infty}\frac{1}{z}dz$
If we compare it with $p-$test for Infinite limits , here $p=1$ ,$a>0$ , It matches with the 2nd one .
So this improper integral also diverges.
Option 4:-
$\large 2*\int_{0}^{1}\frac{1-\cos x}{x^{5}}dx$
$I$=$\large 2*\int_{0}^{1}\frac{1}{x^{5}}dx-2\int_{0}^{1}\frac{\cos x}{x^{5}}dx$
$I$=$I_{1}-I_{2}$
where $I_{1}=$$\large 2*\int_{0}^{1}\frac{1}{x^{5}}dx$ ,$I_{2}=$ $\large 2\int_{0}^{1}\frac{\cos x}{x^{5}}dx$
Now , $\large 2*\int_{0}^{1}\frac{1}{x^{5}}dx$ this integral diverges as it is matches $p-$test for discontinuity case 2 as here p=5 >1 and a=1>0 .
So as $I_{1}$ diverges the whole improper integral $I$ also diverges . So
This is also diverges .
Only correct option is (B) It converges .
Reference