Archive
2nd of March 2017
Mathematics
Q:

Evaluate integral subscript pi divided by 6 end subscript superscript pi divided by 3 end superscript fraction numerator s i n space x space plus space cos x over denominator square root of sin space 2 x end root end fraction space d x

Ahuja8087 Ahuja - CBSE - Class XII Science

Thursday, March 02, 2017 at 16:39:PM

A:

begin mathsize 16px style I equals integral subscript straight pi divided by 6 end subscript superscript straight pi divided by 3 end superscript fraction numerator sin space x plus cos space x over denominator square root of sin space 2 x end root end fraction d x equals integral subscript straight pi divided by 6 end subscript superscript straight pi divided by 3 end superscript fraction numerator sin space x plus cos space x over denominator square root of 1 minus left parenthesis cos space x minus sin space x right parenthesis squared end root end fraction d x
L e t space t equals left parenthesis cos space x minus sin space x right parenthesis d x
rightwards double arrow d t equals d left parenthesis cos space x minus sin space x right parenthesis
rightwards double arrow negative left parenthesis sin x plus cos x right parenthesis d x equals d t
A l s o comma space x equals straight pi over 6 rightwards double arrow t equals cos straight pi over 6 minus sin straight pi over 6 equals fraction numerator square root of 3 minus 1 over denominator 2 end fraction
a n d comma space x equals straight pi over 3 rightwards double arrow t equals cos straight pi over 3 minus sin straight pi over 3 equals fraction numerator 1 minus square root of 3 over denominator 2 end fraction
I equals integral subscript fraction numerator square root of 3 minus 1 over denominator 2 end fraction end subscript superscript fraction numerator 1 minus square root of 3 over denominator 2 end fraction end superscript fraction numerator 1 over denominator square root of 1 minus t squared end root end fraction d t
equals negative open square brackets sin to the power of negative 1 end exponent t close square brackets subscript fraction numerator square root of 3 minus 1 over denominator 2 end fraction end subscript superscript fraction numerator 1 minus square root of 3 over denominator 2 end fraction end superscript
equals negative open square brackets sin to the power of negative 1 end exponent fraction numerator 1 minus square root of 3 over denominator 2 end fraction minus sin to the power of negative 1 end exponent fraction numerator square root of 3 minus 1 over denominator 2 end fraction close square brackets
equals negative sin to the power of negative 1 end exponent fraction numerator 1 minus square root of 3 over denominator 2 end fraction plus sin to the power of negative 1 end exponent fraction numerator square root of 3 minus 1 over denominator 2 end fraction
equals 2 sin to the power of negative 1 end exponent fraction numerator square root of 3 minus 1 over denominator 2 end fraction space space space space space space space space space.... open square brackets sin to the power of negative 1 end exponent left parenthesis negative x right parenthesis equals negative sin to the power of negative 1 end exponent x close square brackets end style

Thursday, March 02, 2017 at 17:46:PM