The integral _ 12 ^ 4 8 , ,2x ( ,x + ,x ) ^3 ,dx equals - Vidyadip

MCQ

The integral $\int_{\frac{\pi }{{12}}}^{\frac{\pi }{4}} {\frac{{8\,\cos \,2x}}{{{{\left( {\tan \,x + \cot \,x} \right)}^3}}}\,dx} $ equals

✓
$\frac{{15}}{{128}}$
B
$\frac{{15}}{{64}}$
C
$\frac{{13}}{{32}}$
D
$\frac{{15}}{{256}}$

✓

Answer

Correct option: A.

$\frac{{15}}{{128}}$

a
$\int\limits_{\frac{\pi }{{12}}}^{\frac{\pi }{4}} {\frac{{\cos 2x}}{{{{\left( {\frac{1}{{\sin 2x}}} \right)}^3}}} = } \int\limits_{\frac{\pi }{{12}}}^{\frac{\pi }{4}} {\cos 2x} \times \sin 2x \cdot {\sin ^2}(2x)dx$

$ = \frac{1}{4}\int\limits_{\frac{\pi }{{12}}}^{\frac{\pi }{4}} {\sin 4x.\left( {1 - \cos 4x} \right)} dx$

$ = \frac{1}{4}\left[ {\int\limits_{\frac{\pi }{{12}}}^{\frac{\pi }{4}} {\sin 4x} - \frac{1}{2}\int\limits_{\frac{\pi }{{12}}}^{\frac{\pi }{4}} {\sin 8x} } \right]$

$ = \frac{1}{4}\left[ {\frac{{\cos 4x}}{4} + \frac{{\cos 8x}}{{16}}} \right]_{\pi /12}^{\pi /4} - \frac{1}{4}\left[ {\frac{{15}}{{23}}} \right]$

$ = \frac{{15}}{{128}}$

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