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$integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals$

$2 \log 2$
$\frac{π}{2} \log 2$
$\log \sqrt{2}$
$\log 2$

hint Hint:

We are aware that differentiation is the process of discovering a function's derivative and integration is the process of discovering a function's antiderivative. Thus, both processes are the antithesis of one another. Therefore, we can say that differentiation is the process of differentiation and integration is the reverse. The anti-differentiation is another name for the integration.
Here we have given: $integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x$ and we have to integrate it. We will use the formula to find the answer.

The correct answer is: $log invisible function application square root of 2$

Now we have given the function as $integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x$ . Here the lower limit is $straight pi over 4$ and upper limit is $straight pi over 2$ . We know that there are some integrals of special functions which makes problem to solve easily. We will use one of them.
We will use substitution method.
So as per the question, we have:
$integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x W e space k n o w space t h a t space c o t x space equals space sin x divided by cos x comma space s o space a p p l y i n g space t h i s comma space w e space g e t colon integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript fraction numerator cos x over denominator sin x end fraction d x H e r e space w e space w i l l space u s e space t h e space f o r m u l a colon integral fraction numerator space space f apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis space end fraction d x equals log ∣ f left parenthesis x right parenthesis ∣ integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript fraction numerator cos x over denominator sin x end fraction d x space equals space ln open vertical bar sin space x close vertical bar subscript begin inline style bevelled straight pi over 4 end style end subscript superscript bevelled straight pi over 2 end superscript E x p a n d i n g space t h i s comma space w e space g e t colon integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals space ln left parenthesis sin space straight pi over 2 right parenthesis minus ln left parenthesis sin space straight pi over 4 right parenthesis space integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals ln left parenthesis 1 right parenthesis minus ln left parenthesis fraction numerator 1 over denominator square root of 2 end fraction right parenthesis integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals 1 half ln left parenthesis 2 right parenthesis$ .

So here we used the concept of integrals of special functions and simplified it. We can also solve it manually but it will take lot of time to come to final answer hence we used the special case and the formula of that. The integral of the given function is $integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript cot invisible function application x d x equals 1 half ln left parenthesis 2 right parenthesis$ .

$integral subscript 0 superscript 1 open parentheses 1 plus e to the power of negative x end exponent close parentheses d x equals$

= $integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x end exponent right parenthesis d x$
>>>Integration of 1 becomes x and integration of e-x becomes -e-x.
>>> = $left parenthesis 1 minus 1 over e plus 1 right parenthesis$
= (2- $1 over e$ )

$integral subscript 0 superscript 1 open parentheses 1 plus e to the power of negative x end exponent close parentheses d x equals$

Maths-General

General

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$integral subscript 0 superscript pi cos invisible function application 3 x d x$

So here we used the concept of integrals of special functions and simplified it. We can also solve it manually but it will take lot of time to come to the final answer hence we will use trigonometric formulas. The integral of the given function is $integral subscript 0 superscript straight pi cos invisible function application 3 xdx equals 0$

$integral subscript 0 superscript pi cos invisible function application 3 x d x$

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The correct answer is:

Now we have given the function as . Here the lower limit is and upper limit is . We know that there are some integrals of special functions which makes problem to solve easily. We will use one of them.We will use substitution method.So as per the question, we have:.

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Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives

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The correct answer is: $log invisible function application square root of 2$

Which of the following pairs is / are correctly matched?

Select the correct answer using the codes given below:

Which of the following pairs is / are correctly matched?

Select the correct answer using the codes given below:

The following kinetic data are provided for a reaction between A and B:

The value of the rate constant for the above reaction is equal to

The following kinetic data are provided for a reaction between A and B:

The value of the rate constant for the above reaction is equal to

The following data are for the decomposition of ammonium nitrite in aqueous solution

The order of the reaction is

The following data are for the decomposition of ammonium nitrite in aqueous solution

The order of the reaction is

The potential energy diagram for a reaction R → P is given below:

$increment H to the power of 0 end exponent$ of the reaction corresponds to the energy

The potential energy diagram for a reaction R → P is given below:

$increment H to the power of 0 end exponent$ of the reaction corresponds to the energy

A graph plotted between $l o g blank k$ vs 1/T for calculating activation energy is shown by

A graph plotted between $l o g blank k$ vs 1/T for calculating activation energy is shown by

In the following groups:
I) OAc,
II) - OMe,
III) -OSO₂Me ,
IV) -OSO₂CF₃ the order of leaving group ability is

In the following groups:
I) OAc,
II) - OMe,
III) -OSO₂Me ,
IV) -OSO₂CF₃ the order of leaving group ability is