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Integrals

Short Answer Type

151.

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

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152.

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

106 Views

153.

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

94 Views

154.
Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction dx space equals space straight pi over 4$

Let I = $integral subscript 0 superscript straight pi divided by 2 end superscript fraction numerator cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction dx$ ...(1)
Then $straight I space equals space integral subscript 0 superscript straight pi divided by 2 end superscript fraction numerator cos to the power of 5 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses over denominator sin to the power of 5 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses plus cos to the power of 5 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction dx space space open square brackets because integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
$therefore space space space straight I space equals space integral subscript 0 superscript straight pi divided by 2 end superscript fraction numerator sin to the power of 5 straight x over denominator cos to the power of 5 straight x plus sin to the power of 5 straight x end fraction dx$ ...(2)
Adding (1) and (2) we get,
$2 space straight I space equals space integral subscript 0 superscript straight pi divided by 2 end superscript open square brackets fraction numerator cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction plus fraction numerator sin to the power of 5 straight x over denominator cos to the power of 5 straight x plus sin to the power of 5 straight x end fraction close square brackets dx space space space space space space equals space integral subscript 0 superscript straight pi divided by 2 end superscript fraction numerator sin to the power of 5 straight x plus cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction dx space equals space integral subscript 0 superscript straight pi divided by 2 end superscript space 1 space dx space equals space open square brackets straight x close square brackets subscript 0 superscript straight pi divided by 2 end superscript space equals space straight pi over 2 minus 0 space equals space straight pi over 2 therefore space space 2 space straight I space equals space fraction numerator straight pi over denominator 2 space end fraction space space space space space space space rightwards double arrow space space space space space space space space straight I space equals space straight pi over 4$

89 Views

155.

Evaluate:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin to the power of begin display style 3 over 2 end style end exponent straight x over denominator sin to the power of begin display style 3 over 2 end style end exponent straight x plus cos to the power of begin display style 3 over 2 end style end exponent straight x end fraction dx space$

108 Views

156.

Evaluate:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin to the power of begin display style 5 over 3 end style end exponent straight x over denominator sin to the power of begin display style 5 over 3 end style end exponent straight x plus cos to the power of begin display style 5 over 3 end style end exponent straight x end fraction dx space equals space straight pi over 4$

94 Views

157.

Evaluate:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of cotx over denominator square root of tanx plus square root of cotx end fraction dx space equals space straight pi over 4$

106 Views

158.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus cot space straight x end fraction space equals space straight pi over 4$

94 Views

Long Answer Type

159.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus square root of tan space straight x end root end fraction space equals space straight pi over 4$

78 Views

Short Answer Type

160.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus square root of cot space straight x end root end fraction space equals space straight pi over 4$