Evaluate by substitution and using properties of definite integral
 
 
integral subscript 0 superscript 1 space log left parenthesis square root of 1 plus x end root space space plus space square root of 1 minus x space end root space right parenthesis space d x

Asked by ahuja8087 | 11th Nov, 2016, 06:02: PM

Expert Answer:

begin mathsize 16px style Consider comma space open parentheses square root of 1 plus straight x end root plus square root of 1 minus straight x end root close parentheses squared space equals space 1 plus straight x plus 1 minus straight x minus 2 square root of 1 plus straight x end root square root of 1 minus straight x end root space equals space 2 open parentheses 1 plus square root of 1 minus straight x squared end root close parentheses
So comma space log open parentheses square root of 1 plus straight x end root plus square root of 1 minus straight x end root close parentheses space equals log open parentheses 2 open parentheses 1 plus square root of 1 minus straight x squared end root close parentheses close parentheses to the power of 1 divided by 2 end exponent equals 1 half open square brackets log 2 plus log space open parentheses 1 plus square root of 1 minus straight x squared end root close parentheses close square brackets
Now comma
integral subscript 0 superscript 1 log open parentheses square root of 1 plus straight x end root plus square root of 1 minus straight x end root close parentheses space dx space
equals space integral subscript 0 superscript 1 open curly brackets 1 half open square brackets log 2 plus log space open parentheses 1 plus square root of 1 minus straight x squared end root close parentheses close square brackets close curly brackets space dx
equals 1 half integral subscript 0 superscript 1 log 2 space dx plus 1 half integral subscript 0 superscript 1 log space open parentheses 1 plus square root of 1 minus straight x squared end root close parentheses space dx end style
 
 

Answered by Rebecca Fernandes | 12th Nov, 2016, 10:31: AM