What is $I_{1}$ equal to ?

\frac{π}{24}

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\frac{π}{18}

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\frac{π}{12}

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\frac{π}{6}

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Solution

The correct option is C $\frac{π}{12}$
$I_{1} = \int_{\frac{π}{6}}^{\frac{π}{3}}$ $\frac{d x}{1 + \sqrt{t a n x}}$
Using the property of definite integrals and substituting x by $x - \frac{π}{6} - \frac{π}{3}$

We get $I_{1} = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{d x}{1 + \sqrt{c o t x}}$
$\Rightarrow I_{1} = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{t a n x} d x}{1 + \sqrt{t a n x}}$
on adding both, we get
$I_{1} + I_{1} = \int_{\frac{π}{6}}^{\frac{π}{3}} d x$
$2 I_{1} = \frac{π}{3} - \frac{π}{6}$
$I_{1} = \frac{π}{12}$

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