Evaluate the following:
$integral subscript 0 superscript 1 cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses dx$

Solution

Let I = $integral subscript 0 superscript 1 cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses dx$
Put x = tan θ so that dx = sec² θ dθ When x = 0, tan θ = 1 ⇒ θ = 0
When $straight x space equals space 1 comma space space space tanθ space equals space 1$    $rightwards double arrow space space space straight theta space equals space straight pi over 4$
$therefore space space space space straight I space equals space space integral subscript 0 superscript straight pi over 4 end superscript cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus tan squared straight theta over denominator 1 plus tan squared straight theta end fraction close parentheses. space sec squared straight theta space dθ space equals space integral subscript 0 superscript straight pi over 4 end superscript cos to the power of negative 1 end exponent left parenthesis cos space 2 straight theta right parenthesis space. sec squared straight theta space dθ$
$equals space integral subscript 0 superscript straight pi over 4 end superscript 2 space straight theta space sec squared straight theta space dθ space equals space 2 integral subscript 0 superscript straight pi over 4 end superscript straight theta. space sec squared straight theta space dθ$
$equals space 2 open curly brackets open square brackets straight theta space tanθ close square brackets subscript 1 superscript straight pi over 4 end superscript minus integral subscript 0 superscript straight pi over 4 end superscript 1. space tanθ space dθ close curly brackets space equals space 2 open curly brackets open square brackets straight theta space tan space straight theta close square brackets subscript 0 superscript straight pi over 4 end superscript plus open square brackets log space cosθ close square brackets subscript 0 superscript straight pi over 4 end superscript close curly brackets$
   $equals space 2 open curly brackets straight pi over 4 space tan straight pi over 4 minus 0 plus log space cos straight pi over 4 minus log space cos space 0 close curly brackets space equals space 2 open square brackets straight pi over 4 cross times 1 minus 0 plus log fraction numerator 1 over denominator square root of 2 end fraction minus log space 1 close square brackets equals space 2 open square brackets straight pi over 4 plus log space 1 space minus space log space square root of 2 close square brackets space equals space straight pi over 2 minus log space 2$

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Integrals

Evaluate the following:
$integral subscript 0 superscript 1 cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses dx$

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