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Integrals

Short Answer Type

151.
By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of sinx over denominator square root of sinx plus square root of cosx end fraction dx$

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

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

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