Prove that:  from Mathematics Integrals

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 Multiple Choice QuestionsShort Answer Type

101.

Evaluate the following integrals
integral subscript negative 1 end subscript superscript 1 space straight x cubed space left parenthesis straight x to the power of 4 plus 1 right parenthesis cubed space dx

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102.

Evaluate the following integrals
integral subscript 0 superscript 1 straight x square root of 1 minus straight x squared end root dx

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103.

Evaluate the following integrals:
integral subscript negative 1 end subscript superscript 1 fraction numerator 5 straight x over denominator left parenthesis 4 plus straight x squared right parenthesis squared end fraction dx



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104.

Evaluate the following integral:
integral subscript 0 superscript 1 fraction numerator 5 straight x over denominator left parenthesis 4 plus straight x squared right parenthesis squared end fraction dx



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 Multiple Choice QuestionsLong Answer Type

105.

Evaluate the following integral:
integral subscript 0 superscript 2 fraction numerator dx over denominator straight x plus 4 minus straight x squared end fraction



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 Multiple Choice QuestionsShort Answer Type

106. Evaluate the following integral:
integral subscript negative 1 end subscript superscript 1 fraction numerator dx over denominator straight x squared plus 2 straight x plus 5 end fraction
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107.

Evaluate integral subscript 0 superscript straight pi over 2 end superscript space square root of sin space straight ϕ end root space cos to the power of 5 straight ϕ space dϕ space

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 Multiple Choice QuestionsLong Answer Type

108.

Evaluate integral subscript 0 superscript 2 fraction numerator 5 straight x plus 1 over denominator straight x squared plus 4 end fraction dx.

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109.

Prove that: integral subscript negative straight a end subscript superscript straight a square root of fraction numerator straight a minus straight x over denominator straight a plus straight x end fraction end root space dx space equals space straight a space straight pi.


Let I = integral subscript negative straight a end subscript superscript straight a space square root of fraction numerator straight a minus straight x over denominator straight a plus straight x end fraction end root dx
Put x = a cos θ , ∴ dx = –a sin θ dθ
When x = a, a = a cos θ ⇒ 1 = cos θ    ⇒ θ = 0
When x = –a, –a = a cos θ ⇒ – 1 = cos θ ⇒ θ = straight pi
therefore                               straight I space equals space integral subscript straight pi superscript 0 square root of fraction numerator straight a minus acosθ over denominator straight a plus acosθ end fraction end root space space space space left parenthesis negative straight a space sin space straight theta space dθ right parenthesis
                         equals space straight a space integral subscript 0 superscript straight pi square root of fraction numerator 1 minus cosθ over denominator 1 plus cosθ end fraction end root. space space sin space straight theta space dθ space equals space straight a space integral subscript 0 superscript straight pi square root of fraction numerator 2 sin squared begin display style straight theta over 2 end style over denominator 2 cos squared begin display style straight theta over 2 end style end fraction end root. space sin space straight theta space dθ
    equals space straight a space integral subscript 0 superscript straight pi fraction numerator sin space begin display style straight theta over 2 end style over denominator cos begin display style straight theta over 2 end style end fraction. space 2 space sin space straight theta over 2 space cos straight theta over 2 space dθ space equals space straight a space integral subscript 0 superscript straight pi space 2 space sin squared straight theta over 2 space dθ
   equals space straight a integral subscript 0 superscript straight pi left parenthesis 1 minus cos space straight theta right parenthesis dθ space equals space straight a space open square brackets straight theta minus sinθ close square brackets subscript 0 superscript straight pi

   equals space straight a open square brackets left parenthesis straight pi minus sin space straight pi right parenthesis space minus space left parenthesis 0 minus sin space 0 right parenthesis close square brackets
equals space straight a open square brackets left parenthesis straight pi minus 0 right parenthesis space minus space left parenthesis 0 minus 0 right parenthesis close square brackets space equals aπ.


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110.

Evaluate  integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x space plus space cos space straight x over denominator square root of sinx space cosx end root end fraction dx.

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