Evaluate   from Mathematics Integrals
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    Integrals

    Multiple Choice QuestionsShort Answer Type

    161.

    Show that:
    integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos space straight x over denominator sin space straight x space plus space cos space straight x end fraction dx


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    Answer

    Multiple Choice QuestionsLong Answer Type

    162.

    Evaluate integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator dx over denominator 1 plus square root of tanx end fraction

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    Answer
    163.

    Evaluate integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction dx

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    Answer

    Multiple Choice QuestionsShort Answer Type

    164.

    Evaluate integral subscript 0 superscript straight pi space sin squared straight x space cos cubed straight x space dx.

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    Answer
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    165.

    Evaluate integral subscript negative straight pi over 4 end subscript superscript straight pi over 4 end superscript space sin squared straight x space dx

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    Answer
    166.

    Evaluate integral subscript negative straight pi over 4 end subscript superscript straight pi over 4 end superscript cos squared xdx

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    Answer
    167.

    By using the properties of definite integrals, evaluate
    integral subscript negative straight pi over 2 end subscript superscript straight pi over 2 end superscript cosx space dx

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    Answer
    168.

    Evaluate integral subscript negative 1 end subscript superscript 1 sin to the power of 5 straight x space cos to the power of 4 straight x space dx

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    Answer
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    169.

    Evaluate integral subscript negative 1 end subscript superscript 1 space straight x to the power of 17 space cos to the power of 4 straight x space dx.

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    Answer
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    170.

    Evaluate  integral subscript negative 1 end subscript superscript 1 space log space open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses dx.

    integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space minus integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space space plus integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx
    rightwards double arrow space space space integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space 0
    Let  straight I space equals space integral subscript negative 1 end subscript superscript 1 space log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses dx
    Let   straight f left parenthesis straight x right parenthesis space equals space log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses
    therefore space space space straight f left parenthesis negative straight x right parenthesis space equals space log open parentheses fraction numerator 2 minus straight x over denominator 2 plus straight x end fraction close parentheses space equals space log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses to the power of negative 1 end exponent space equals space minus log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses space equals space minus straight f left parenthesis straight x right parenthesis therefore space space space space space space space straight f left parenthesis straight x right parenthesis space is space an space odd space function space of space straight x
    therefore space space space straight I space equals space 0 space space space space space space space space space space space space open square brackets because space space integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space 0 space space if space straight f left parenthesis straight x right parenthesis space is space an space odd space function space of space straight x close square brackets
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