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integral subscript 0 superscript pi divided by 2 end superscript   fraction numerator 1 over denominator 4 plus 5 cos invisible function application x end fraction d x equals

  1. 1 fifth log invisible function application 2
  2. 1 half log invisible function application 2
  3. 1 third log invisible function application 3
  4. 1 third log invisible function application 2

hintHint:

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 0 superscript pi divided by 2 end superscript   fraction numerator 1 over denominator 4 plus 5 cos invisible function application x end fraction d x and we have to integrate it. We will use the formula to find the answer.

The correct answer is: 1 third log invisible function application 2


    Now we have given the function as integral subscript 0 superscript pi divided by 2 end superscript   fraction numerator 1 over denominator 4 plus 5 cos invisible function application x end fraction d x. Here the lower limit is 0 and upper limit is straight pi over 2. We know that there are some integrals where we can use substitution method which makes problem to solve easily. We will use one of them.
    We will use the substitution method.
    So as per the question, we have:
    integral subscript 0 superscript pi divided by 2 end superscript   fraction numerator 1 over denominator 4 plus 5 cos invisible function application x end fraction d x
L e t space tan space left parenthesis x divided by 2 right parenthesis space equals space t comma space t h e n space d i f f e r e n t i a t i o n space i t comma space w e space g e t colon
d t equals s e c squared left parenthesis x divided by 2 right parenthesis cross times 1 half d x
S i m p l i f y i n g space i t comma space w e space g e t colon
d x equals fraction numerator 2 d t over denominator s e c squared begin display style x over 2 end style end fraction
d x equals fraction numerator 2 d t over denominator 1 plus tan squared begin display style x over 2 end style end fraction equals fraction numerator 2 d t over denominator 1 plus t squared end fraction
C h a n g i n g space t h e space l i m i t s comma space w e space g e t colon
i f space x equals 0 comma space t h e n space t equals 0
i f space x equals straight pi divided by 2 comma space then space straight t equals 1
N o w space w e space k n o w space t h a t colon
cos x equals fraction numerator 1 minus tan squared x over 2 over denominator 1 plus tan squared x over 2 end fraction equals fraction numerator 1 minus t squared over denominator 1 plus t squared end fraction
N o w space u sin g space t h e space o r i g i n a l space i n t e g r a l comma space w e space g e t colon
integral subscript 0 superscript pi divided by 2 end superscript   fraction numerator 1 over denominator 4 plus 5 cos invisible function application x end fraction d x equals integral subscript 0 superscript 1   fraction numerator fraction numerator 2 d t over denominator 1 plus t squared end fraction over denominator 4 plus 5 left parenthesis fraction numerator 1 minus t squared over denominator 1 plus t squared end fraction right parenthesis end fraction
equals 2 integral fraction numerator d t over denominator 4 left parenthesis 1 plus t squared right parenthesis plus 5 left parenthesis 1 minus t squared right parenthesis end fraction
equals 2 integral fraction numerator d t over denominator 4 plus 5 plus 4 t squared minus 5 t squared end fraction
equals 2 integral fraction numerator d t over denominator 9 minus t squared end fraction
equals negative 2 integral fraction numerator d t over denominator t squared minus 3 squared end fraction
U sin g space t h e space f o r m u l a space o f colon
integral fraction numerator 1 over denominator x squared minus a squared end fraction d x equals fraction numerator 1 over denominator 2 a end fraction log open vertical bar fraction numerator x minus a over denominator x plus a end fraction close vertical bar plus c
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
equals negative 2 integral fraction numerator d t over denominator t squared minus 3 squared end fraction equals negative 2 cross times fraction numerator 1 over denominator 2 left parenthesis 3 right parenthesis end fraction cross times log open vertical bar fraction numerator t minus 3 over denominator t plus 3 end fraction close vertical bar plus c
S u b s t i t u t i n g space t h e space l i m i t s comma space w e space g e t colon
equals negative 2 integral fraction numerator d t over denominator t squared minus 3 squared end fraction equals negative 2 cross times fraction numerator 1 over denominator 2 left parenthesis 3 right parenthesis end fraction cross times log open vertical bar fraction numerator 1 minus 3 over denominator 1 plus 3 end fraction close vertical bar plus 2 cross times fraction numerator 1 over denominator 2 left parenthesis 3 right parenthesis end fraction cross times log open vertical bar fraction numerator 0 minus 3 over denominator 0 plus 3 end fraction close vertical bar
fraction numerator 1 over denominator left parenthesis 3 right parenthesis end fraction cross times log open vertical bar fraction numerator negative 2 over denominator 4 end fraction close vertical bar plus fraction numerator 1 over denominator left parenthesis 3 right parenthesis end fraction cross times log open vertical bar fraction numerator negative 3 over denominator 3 end fraction close vertical bar
fraction numerator 1 over denominator left parenthesis 3 right parenthesis end fraction cross times log open vertical bar fraction numerator negative 1 over denominator 2 end fraction close vertical bar plus fraction numerator 1 over denominator left parenthesis 3 right parenthesis end fraction cross times log open vertical bar negative 1 close vertical bar
fraction numerator 1 over denominator left parenthesis 3 right parenthesis end fraction cross times log left parenthesis fraction numerator negative 1 over denominator 2 end fraction cross times negative 1 right parenthesis
fraction numerator 1 over denominator left parenthesis 3 right parenthesis end fraction cross times log left parenthesis 1 half right parenthesis

    So here we used the concept of integrals 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 substitution method which makes problem to solve easily. The integral of the given function is fraction numerator 1 over denominator left parenthesis 3 right parenthesis end fraction cross times log left parenthesis 1 half right parenthesis

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    integral subscript 0 superscript pi divided by 4 end superscript   x sec squared invisible function application x d x equals

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    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 formula. The integral of the given function is integral subscript 0 superscript pi divided by 4 end superscript   x sec squared invisible function application x d x equals straight pi over 4 minus log square root of 2

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    integral subscript negative 1 end subscript superscript 1   x e to the power of x d x equals

    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 formula to solve. The integral of the given function isintegral subscript negative 1 end subscript superscript 1 x e to the power of x d x equals 2 over e

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    integral subscript 0 superscript straight infinity   open parentheses straight a to the power of negative straight x end exponent minus straight b to the power of negative straight x end exponent close parentheses dx equals left parenthesis straight a greater than 1 comma straight b greater than 1 right parenthesis

    Therefore, the integration of integral subscript 0 superscript straight infinity   open parentheses straight a to the power of negative straight x end exponent minus straight b to the power of negative straight x end exponent close parentheses dx equals left parenthesis straight a greater than 1 comma straight b greater than 1 right parenthesis is fraction numerator 1 over denominator log a end fraction minus fraction numerator 1 over denominator log b end fraction.

    integral subscript 0 superscript straight infinity   open parentheses straight a to the power of negative straight x end exponent minus straight b to the power of negative straight x end exponent close parentheses dx equals left parenthesis straight a greater than 1 comma straight b greater than 1 right parenthesis

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    Therefore, the integration of integral subscript 0 superscript straight infinity   open parentheses straight a to the power of negative straight x end exponent minus straight b to the power of negative straight x end exponent close parentheses dx equals left parenthesis straight a greater than 1 comma straight b greater than 1 right parenthesis is fraction numerator 1 over denominator log a end fraction minus fraction numerator 1 over denominator log b end fraction.

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