Why is the following equation not correct? 
Q)Point A moves uniformly with velocity $v$ so that the vector $v$ is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity $u<v$. At the initial moment of time $v⊥u$ and the points are separated by a distance $l$.The time at which the points converge is given as?


Let at time t, the the angle between the $v$ and horizontal be $\theta$. Seeing this I made two equations,
$$uT=\int_0^T ​vcos\theta dt$$ (As the horizontal distance covered has to be equal)
And,
$$l=\int_0^T vsin\theta dt $$because the vertical distance covered by the object A has to be equal to l, since they were initially l distance apart.However, the 2nd equation is wrong, the correct equation being given by:
$$l=\int_0^T (v-ucos\theta) dt$$
I am unable to find see how the relative distance between the the points is equal to l as the l is in different direction from $v-ucos\theta$. Also, why isn't $l=\int_0^T vsin\theta dt $ correct, as the $sin\theta$ component is in vertical direction and what does $v sin\theta$ integration equate to?
 A: It is important to keep in the back of your mind that $\theta$ is really a function of time $\theta(t)$ and that using this function, one can find the velocity (not speed) of particle $A$ at time $t$: $v(\cos \theta(t), \sin \theta(t))$. Ultimately, we can plug and chug with this velocity vector and get a differential equation of $\theta$; however, this equation is complicated, and solving it is unnecessary. Instead, we can manipulate the given information to eliminate $\theta$ from a system of equations and parse out only $T$,  because that is all that this problem (Irodov 1.13) is asking for.
It is immediate that from time $0$ until time $T$, particle $A$ must cover a horizontal distance of $u \cdot T$ and a vertical distance of $l$.
Your integrals:
$$l = \int_0^T v \sin \theta(t) \ \mathrm{d} t \\ u \cdot T = \int_0^T v \cos \theta(t) \ \mathrm{d} t$$
are indeed correct, but these equations cannot be manipulated to get rid of the sines nor the cosines. We want two integrals both containing $\sin \theta(t)$ or both containing $\cos \theta(t)$.
We can accomplish this through the integral:
$$\int_0^T v - u \cdot \cos \theta(t) \ \mathrm{d} t$$
The value $v - u \cdot \cos \theta(t)$ has a geometric significance. A picture helps quite a bit here:

The vectors in the image represent the velocities at a certain time. Say, $\vec{u}(0)$ means the vector point along the direction of the movement of particle $B$ with its magnitude (constant $u$) at the time $t=0$. Note that the vector $\vec{v}(t)$ has been copied over to the location of particle $B$, just like in OP's diagram. Let's think about the orange vector parallel to $\vec{v}(t)$ in the diagram.
This orange segment corresponds to the quantity $u \cos \theta(t)$, pointing in the direction of $\hat{v}$, and it represents the component of $B$'s velocity in the direction that $A$ is travelling in. In other words, it's the amount that the total length of $A$'s journey must increase per unit time, due to the movement of particle $B$. In a small amount of time, say $\mathrm{d} t$, $A$'s travel distance increases, by $u \cos \theta(t) \ \mathrm{d} t$.
At the same time, particle $A$ has been steadily gaining on straight-line distance. The premise of the problem is that its velocity is always pointed along the straight line connecting it and particle $B$. Therefore, in a small amount of time, the straight-line distance between the two particles closes, due to the movement of particle $A$ by $v \ \mathrm{d} t$, as its velocity is constant and always pointing at particle $B$.
Then, $A$ gains straight-line distance on $B$, in this small period of time, the exact amount being:
$$
\Delta \mathrm{Straight \ Line \ Distance} = (v - u \cdot \cos \theta(t)) \ \mathrm{d} t
$$
Over the timespan ranging from $0$ to $T$, $A$ must gain a total of $l$ straight-line distance on $B$. We can find this total value by simply integrating over time:
$$l = \int_0^T v - u \cdot \cos \theta(t) \ \mathrm{d} t$$
We can then rearrange this integral (keep in mind that $v$ is a constant):
$$ \int_0^T  \cos \theta(t) \ \mathrm{d} t = \frac{1}{u}(vT - l)$$
but on the other hand, the other integral from earlier, telling us about the horizontal distance covered in this timespan tells us:
$$\int_0^T \cos \theta(t) \ \mathrm{d} t = \frac{uT}{v}$$
Equate the two to find that:
$$
T = \frac{v}{v^2 - u^2} l
$$
