If $x = \mathop \smallint \limits_2^{\sin t} {\sin ^{ - 1}}\,\,\theta \,d\theta $ and $y = \mathop \smallint \limits_n^{\sqrt t } \,\,\frac{{\sin \,\,{\theta ^2}}}{\theta }\,\,d\;\theta $ then $\frac{{dy}}{{dx}}$ is equal to

$\frac{{2\,{t^2}}}{{\tan \,\,t}}$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

x = \int \limits_2^{\sin t} {\sin ^{ - 1}}\,\,\theta \,d\theta

and

y = \int \limits_n^{\sqrt t } \,\,\frac{{\sin \,\,{\theta ^2}}}{\theta }\,\,d\;\theta

then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: A

\frac{{\tan \,\,t}}{{2\,{t^2}}}

Option A (

\frac{{\tan t}}{{2{t^2}}}

) is correct because it is the derivative dy/dx found using Leibniz's rule and the chain rule.