How do you integrate $\int{dx\dfrac{\cos x}{\sqrt{1-2\sin x}}}$ ?

Hint: We are given a function as $\int{dx\dfrac{\cos x}{\sqrt{1-2\sin x}}}$, we have to integrate this function, we start by learning about integral, we will be needing the knowledge that how are trigonometry ratio connected to each other. We will use the substitution method we substitute $1-2\sin x$ as ‘t’ then we will use $\int{{{x}^{n}}dn=\dfrac{{{x}^{n+1}}}{n+1}}$ . We will use these to form our solution.

Complete step by step answer:
We are given a function as $\int{dx\dfrac{\cos x}{\sqrt{1-2\sin x}}}$. We have to integrate them, integration is a method of adding the slices to find the full. Now we have to integrate $dx\dfrac{\cos x}{\sqrt{1-2\sin x}}$ .
As we know that sin x and cos x are related to each other as they are derived from one another. So, using this knowledge we will use the substitution method to change one of the things.
We consider $1-2\sin x$ as ‘t’.
So, $1-2\sin x=t$
Now we differentiate both sides, we get –
$0-2\cos xdx=dt$
Simplifying, we get –
$\cos xdx=\dfrac{dt}{-2}$ ………………………… (1)
We will use this to change our function in terms of ‘t’.
So, $\int{\dfrac{\cos x}{\sqrt{1-2\sin x}}}=\int{\dfrac{1}{\sqrt{t}}\dfrac{dt}{\left( -2 \right)}}$ using (1)
Taking constant (-2) our, we get –
$=\dfrac{1}{-2}\int{\dfrac{dt}{\sqrt{t}}}$
As $\dfrac{1}{\sqrt{t}}={{t}^{{}^{-1}/{}_{2}}}$
So, $=-\dfrac{1}{2}\int{{{t}^{{}^{-1}/{}_{2}}}dt}$
Now we will use $\int{{{x}^{n}}dn=\dfrac{{{x}^{n+1}}}{n+1}}$
So for $\int{{{t}^{{}^{-1}/{}_{2}dt}}}$ , we have $n=-\dfrac{1}{2}$ .
So, our integral become
$=-\dfrac{1}{2}\left( \dfrac{{{t}^{{}^{-1}/{}_{2}+1}}}{-\dfrac{1}{2}+1} \right)+c$
By simplifying, we get –
$=-\dfrac{1}{2}\left( \dfrac{{{t}^{{}^{1}/{}_{2}}}}{\dfrac{1}{2}} \right)+c$
So we get –
$=-{{t}^{{}^{1}/{}_{2}}}$ as ${{t}^{{}^{1}/{}_{2}}}=\sqrt{t}$
So, $=-\sqrt{t}+c$
Now we replace ‘t’ as $1-2\sin x$ back into its value. So, we get –
$\int{\dfrac{\cos x}{\sqrt{1-2\sin x}}}=-\sqrt{1-2\sin x}+c$

Note: Remember that in case of indefinite integral, that is integral without limit it is necessary to replace or substitute back the original value of the function back which we replaced earlier.
While in case of definite integral no such thing is necessary as we change function. We also change the limit story with it. So, we did not require changing back the function.
We simply apply the limit there but yes with change in function, the limit must also be changed accordingly.