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Integrals

Question
CBSEENMA12036053

The Integral integral subscript 0 superscript straight pi square root of 1 plus 4 space sin squared straight x over 2 minus 4 sin straight x over 2 dx end root is equal to

  • π-4

  • fraction numerator 2 space straight pi over denominator 3 end fraction minus 4 minus 4 square root of 3
  • 4 square root of 3 minus 4
  • 4 square root of 3 minus 4 minus 4 square root of 3

Solution

D.

4 square root of 3 minus 4 minus 4 square root of 3

By using the formula, vertical line straight x minus straight a vertical line space equals open curly brackets table attributes columnalign left end attributes row cell straight x minus straight a comma space space space space space space space space space straight x greater or equal than straight a end cell row cell negative left parenthesis straight x minus straight a right parenthesis comma space space space straight x less than straight a end cell end table close
It breaks given integral in two parts and then integrates separately.
integral subscript 0 superscript straight x square root of open parentheses 1 minus 2 sin straight x over 2 close parentheses squared end root dx space equals space integral subscript 0 superscript straight pi vertical line 1 minus 2 space sin straight x over 2 vertical line dx
equals integral subscript 0 superscript straight pi over 3 end superscript open parentheses 1 minus 2 space sin space straight x over 2 close parentheses dx space minus integral subscript straight pi over 3 end subscript superscript straight pi open parentheses 1 minus 2 space sin space straight x over 2 close parentheses dx
equals space open parentheses straight x plus 4 space cos space straight x over 2 close parentheses subscript 0 superscript straight pi over 3 end superscript space minus space open parentheses straight x plus 4 space cos space straight x over 2 close parentheses subscript straight pi over 3 end subscript superscript straight pi
equals space 4 square root of 3 space minus 4 minus straight pi over 3

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