Answer

Verified

341.7k+ views

**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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Why should electric field lines never cross each other class 12 physics CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Write the difference between soap and detergent class 10 chemistry CBSE

Give 10 examples of unisexual and bisexual flowers

Differentiate between calcination and roasting class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the difference between anaerobic aerobic respiration class 10 biology CBSE

a Why did Mendel choose pea plants for his experiments class 10 biology CBSE