Evaluate the following integral:
$integral subscript 0 superscript 1 fraction numerator left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis squared over denominator 1 plus straight x squared end fraction dx$

Solution

Let I = $integral subscript 0 superscript 1 fraction numerator left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis squared over denominator 1 plus straight x squared end fraction dx$
Put $tan to the power of negative 1 end exponent straight x space equals space straight y comma space space space space space therefore space space space space fraction numerator 1 over denominator 1 plus straight x squared end fraction dx space equals space dy$
When x = 0, $straight y space equals tan to the power of negative 1 end exponent 0 space equals space 0$
When x = 1, $straight y space equals space tan to the power of negative 1 end exponent space 1 space space equals space straight pi over 4$
$therefore$ $straight I space equals space integral subscript 0 superscript straight pi over 4 end superscript space straight y squared dy space equals space open square brackets straight y cubed over 3 close square brackets subscript 0 superscript straight pi over 4 end superscript space equals space 1 third open square brackets straight pi cubed over 64 minus 0 close square brackets space equals space straight pi cubed over 192$

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Integrals

Evaluate the following integral:
$integral subscript 0 superscript 1 fraction numerator left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis squared over denominator 1 plus straight x squared end fraction dx$

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