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Integrals

Multiple Choice Questions

141.

If $straight f left parenthesis straight x right parenthesis space equals space integral subscript 0 superscript straight x space straight t space sin space straight t space dt comma space space then space straight f apostrophe left parenthesis straight x right parenthesis space is$

cosx + x sin x
x sinx
x cosx
x cosx

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Short Answer Type

142.

Evaluate $integral subscript 0 superscript straight pi divided by 2 end superscript space sin squared straight x space dx$

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143. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi over 2 end superscript space cos squared straight x space dx

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144. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction dx

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145. By using the properties of definite integrals, evaluate the following:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of cosx over denominator square root of cosx plus square root of sinx end fraction dx$

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of cos space straight x end root over denominator square root of cos space straight x space end root plus square root of sin space straight x end root end fraction dx$ ...(1)
Then $straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root over denominator square root of cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses plus square root of sin space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root end root end fraction dx$
   $open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
$therefore space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction dx$ ...(2)
Adding (1) and (2), we get,
   $2 space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript space open square brackets fraction numerator square root of cos space straight x end root over denominator square root of cos space straight x end root plus square root of sin space straight x end root end fraction plus fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction close square brackets dx space space space space space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of cos space straight x end root plus square root of sin space straight x end root over denominator square root of cos space straight x end root plus square root of sin space straight x end root end fraction dx space equals space integral subscript 0 superscript straight pi over 2 end superscript space 1. space dx space space space space space space equals space open square brackets straight x close square brackets subscript 0 superscript straight pi over 2 end superscript space equals space straight pi over 2 minus 0$
$therefore space space space 2 space space straight I space space equals space straight pi over 2$    $rightwards double arrow space space space straight I space equals space straight pi over 4$

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Long Answer Type

146.

Prove that: $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus tanx end fraction space equals space straight pi over 4.$

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147.

Evaluate the following:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus tan cubed straight x end fraction.$

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148.

Evaluate: $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus tan to the power of 5 straight x end fraction dx$

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149.

Evaluate $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus cot cubed straight x end fraction dx$

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Short Answer Type

150.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x over denominator sin space straight x plus space cos space straight x end fraction dx space equals space straight pi over 4$

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