If $\mathop \smallint \limits_{\pi /2}^x \,\,\sqrt {3\, - \,2\,\,{{\sin }^2}\,\,u} \,du\,\, + \,\,\mathop \smallint \nolimits_0^y \,\,\cos \,\,t\,dt = 0$ then, $\frac{{dy}}{{dx}}$ is equal to

$\frac{{\sqrt {4 - 3\,\,{{\sin }^2}\,\,x} }}{{\cos \,\,y}}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Hard

1 of 1

Application +4 | -1

\int \limits_{\pi /2}^x \,\,\sqrt {3\, - \,2\,\,{{\sin }^2}\,\,u} \,du\,\, + \,\,\int \nolimits_0^y \,\,\cos \,\,t\,dt = 0

then,

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

is equal to

View answer and explanation

Correct option: B

- \,\,\frac{{\sqrt {3\,\, - \,\,2\,\,{{\sin }^2}\,\,x} }}{{\cos \,\,y}}

Option B is correct because differentiating the given equation with respect to x and solving for dy/dx yields the stated result.

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