Evaluate  from Mathematics Integrals

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 Multiple Choice QuestionsMultiple 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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 Multiple Choice QuestionsShort 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

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 Multiple Choice QuestionsLong 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


Let I = integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus cot cubed straight x end fraction dx space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus begin display style fraction numerator cos cubed straight x over denominator sin cubed straight x end fraction end style end fraction dx
therefore space space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin cubed straight x over denominator sin cubed straight x plus cos cubed straight x end fraction dx                     ...(1)

rightwards double arrow space space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin cubed open parentheses begin display style straight pi over 2 end style minus straight x close parentheses over denominator sin cubed open parentheses begin display style straight pi over 2 end style minus straight x close parentheses plus cos cubed open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction dx

therefore space space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos cubed straight x over denominator cos cubed straight x plus sin cubed straight x 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 fraction numerator sin cubed straight x plus cos cubed straight x over denominator sin cubed straight x plus cos cubed straight x end fraction dx

therefore                    2 space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript 1 space dx space space space rightwards double arrow space space space space space 2 space straight I space equals space open square brackets straight x close square brackets subscript 0 superscript straight pi over 2 end superscript

therefore space space space space space 2 space straight I space equals space straight pi over 2 minus 0 space space space rightwards double arrow space space straight I space equals space straight pi over 4

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 Multiple Choice QuestionsShort 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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