JEE Mains · Maths · STD 12 - 7.2 definite integral
The integral \(\int_{\frac{\pi }{{12}}}^{\frac{\pi }{4}} {\frac{{8\,\cos \,2x}}{{{{\left( {\tan \,x + \cot \,x} \right)}^3}}}\,dx} \) equals
- A \(\frac{{15}}{{128}}\)
- B \(\frac{{15}}{{64}}\)
- C \(\frac{{13}}{{32}}\)
- D \(\frac{{15}}{{256}}\)
Answer & Solution
Correct Answer
(A) \(\frac{{15}}{{128}}\)
Step-by-step Solution
Detailed explanation
\(\int\limits_{\frac{\pi }{{12}}}^{\frac{\pi }{4}} {\frac{{\cos 2x}}{{{{\left( {\frac{1}{{\sin 2x}}} \right)}^3}}} = } \int\limits_{\frac{\pi }{{12}}}^{\frac{\pi }{4}} {\cos 2x} \times \sin 2x \cdot {\sin ^2}(2x)dx\)…
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