If $\mathop \smallint \nolimits_0^p \,\,\frac{{dx}}{{1\,\, + \,\,4{x^2}}} = \frac{\pi }{8}$, then the value of $p$ is

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \nolimits_0^p \,\,\frac{{dx}}{{1\,\, + \,\,4{x^2}}} = \frac{\pi }{8}

, then the value of

p

View answer and explanation

Correct option: D

\frac{1}{2}

Option D (

\frac{1}{2}

) is correct because solving the definite integral and equating it to

\frac{\pi}{8}

yields

p = \frac{1}{2}

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