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