Evaluate the definite integral:
$integral subscript 0 superscript 1 divided by square root of 2 end superscript fraction numerator sin to the power of negative 1 end exponent straight x over denominator left parenthesis 1 minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction dx$

Solution

Let I = $integral subscript 0 superscript 1 divided by square root of 2 end superscript fraction numerator sin to the power of negative 1 end exponent straight x over denominator left parenthesis 1 minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction dx$
Put $straight x space equals space sin space straight theta space space space or space space straight theta space equals space sin to the power of negative 1 end exponent straight x comma space space space space therefore space space space dx space equals space space cosθ space dθ$
When x = 0, $sin space straight theta space equals space 0 space space space space space space space space space space rightwards double arrow space space space space space space straight theta space space equals 0$
When $straight x space equals space 1 half comma space space sin space straight theta space equals space fraction numerator 1 over denominator square root of 2 end fraction space space space space space space space rightwards double arrow space space space space straight theta space equals space straight pi over 4$
$therefore space space space space straight I space equals space integral subscript 0 superscript straight pi over 4 end superscript fraction numerator straight theta over denominator left parenthesis 1 minus sin squared straight theta right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction cos space straight theta space dθ space equals space integral subscript 0 superscript straight pi over 4 end superscript fraction numerator straight theta space cosθ over denominator cos cubed straight theta space end fraction dθ space equals space integral subscript 0 superscript straight pi over 4 end superscript fraction numerator straight theta over denominator cos squared straight theta end fraction dθ$
$equals space integral subscript 0 superscript straight pi over 4 end superscript space straight theta space sec squared space straight theta space dθ space equals space open square brackets straight theta space tan space straight theta close square brackets subscript 0 superscript straight pi over 4 end superscript space minus space integral subscript 0 superscript straight pi over 4 end superscript 1. space tan space straight theta space dθ equals space open square brackets straight theta space tan space straight theta close square brackets subscript 0 superscript straight pi over 4 end superscript plus open square brackets log space open vertical bar cos space straight theta close vertical bar close square brackets subscript 0 superscript straight pi over 4 end superscript equals space open parentheses straight pi over 4 tan straight pi over 4 minus 0 close parentheses space plus space open square brackets log space open vertical bar cos space straight pi over 4 close vertical bar minus log space left parenthesis cos space 0 right parenthesis close square brackets equals space straight pi over 4 cross times 1 space minus space 0 space plus space log space fraction numerator 1 over denominator square root of 2 end fraction minus space log space 1 space equals space straight pi over 4 plus log space 1 space minus space log space 2 to the power of 1 half end exponent minus 0 space equals space straight pi over 4 minus 1 half log space 2$

For Daily Free Study Material Join wiredfaculty

Integrals

Some More Questions From Integrals Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction