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Integrals

Question
CBSEENMA12036127

The integral  integral subscript straight pi over 4 end subscript superscript fraction numerator 3 straight pi over denominator 4 end fraction end superscript fraction numerator dx over denominator 1 plus space cos space straight x end fractionis equal to

  • -1

  • -2

  • 2

  • 4

Solution

C.

2

straight I space equals integral subscript straight pi over 4 end subscript superscript fraction numerator 3 straight pi over denominator 4 end fraction end superscript fraction numerator dx over denominator 1 plus space cos space straight x end fraction space space..... space left parenthesis 1 right parenthesis
I space equals integral subscript straight pi over 4 end subscript superscript fraction numerator 3 straight pi over denominator 4 end fraction end superscript fraction numerator dx over denominator 1 minus space cos space straight x end fraction space space..... space left parenthesis 2 right parenthesis
Adding space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis
2 straight I space equals integral subscript straight pi over 4 end subscript superscript fraction numerator 3 straight pi over denominator 4 end fraction end superscript fraction numerator 2 over denominator sin squared space straight x end fraction dx
straight I space equals integral subscript straight pi over 4 end subscript superscript fraction numerator 3 straight pi over denominator 4 end fraction end superscript space cosec squared space straight x space dx
straight I space equals space minus space left parenthesis cot space straight x space right parenthesis subscript straight pi divided by 4 end subscript superscript 3 straight pi divided by 4 end superscript space equals 2 space space

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