Evaluate the following:
$integral subscript 0 superscript 1 tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction close parentheses dx$

Solution

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

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Integrals

Evaluate the following:
$integral subscript 0 superscript 1 tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction close parentheses dx$

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