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Evaluate: $integral subscript 0 superscript straight pi over 4 end superscript space secx square root of fraction numerator 1 minus sin space straight x over denominator 1 plus sin space straight x end fraction end root dx.$

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112.

Evaluate: $integral subscript 0 superscript straight a space sin to the power of negative 1 end exponent space open parentheses square root of fraction numerator straight x over denominator straight a plus straight x end fraction end root close parentheses dx.$

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Short Answer Type

113.

Prove that:

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Long Answer Type

114.

Evaluate:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 4 cosx plus 2 sinx end fraction.$

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115.

Evaluate
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 2 space cosx space plus space 4 space sinx end fraction$

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116.

Evaluate
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator sinx plus square root of 3 cosx end fraction dx$

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Short Answer Type

117.

Prove that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 9 plus 16 space cos squared straight x end fraction space equals space straight pi over 30$

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118.

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

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119.
Prove that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space 2 straight ϕ space dϕ over denominator sin to the power of 4 straight ϕ plus cos to the power of 4 straight ϕ end fraction space equals space straight pi over 2$

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space 2 straight ϕ over denominator sin to the power of 4 straight ϕ plus cos to the power of 4 straight ϕ end fraction dϕ space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 2 sinϕcosϕ over denominator sin to the power of 4 straight ϕ plus cos to the power of 4 straight ϕ end fraction dϕ$

$equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator begin display style fraction numerator 2 space sin space straight ϕ space cosϕ over denominator cos to the power of 4 straight ϕ end fraction end style over denominator begin display style fraction numerator sin to the power of 4 straight ϕ over denominator cos to the power of 4 straight ϕ end fraction end style plus begin display style fraction numerator cos to the power of 4 straight ϕ over denominator cos to the power of 4 straight ϕ end fraction end style end fraction dϕ space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 2 space tanϕ space sec squared space straight ϕ space dϕ over denominator tan to the power of 4 straight ϕ plus 1 end fraction$
Put tan² ϕ = y , ∴ 2 tan ϕ sec² ϕ dϕ = dy When ϕ = 0, y = tan² 0 = 0
When $straight ϕ space equals space straight pi over 2 comma space space straight y space equals tan squared straight pi over 2 space rightwards arrow space space infinity$

$therefore space space space space space straight I space equals space integral subscript 0 superscript infinity fraction numerator dy over denominator straight y squared plus 1 end fraction space equals space open square brackets tan to the power of negative 1 end exponent space straight y close square brackets subscript 0 superscript infinity space equals space tan to the power of negative 1 end exponent infinity space minus space tan to the power of negative 1 end exponent 0 space equals space straight pi over 2 minus 0 space equals space straight pi over 2$

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120.

Prove that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cosx space dx over denominator left parenthesis 1 plus sin space straight x right parenthesis space left parenthesis 2 plus sinx right parenthesis end fraction space equals space log space 4 over 3$