JEE Mains · Maths · STD 12 - 7.1 indefinite integral
\(\int {\frac{{{{\sin }^8}\,x - {{\cos }^8}\,x}}{{\left( {1 - 2\,{{\sin }^2}\,x\,{{\cos }^2}\,x} \right)}}} dx\) is equal to
- A \(\frac{1}{2}\,\sin \,2x + c\)
- B \(-\frac{1}{2}\,\sin \,2x + c\)
- C \(-\frac{1}{2}\,\sin \,x + c\)
- D \( - \sin {^2}\,x + c\)
Answer & Solution
Correct Answer
(B) \(-\frac{1}{2}\,\sin \,2x + c\)
Step-by-step Solution
Detailed explanation
\({\rm{ Let I}} = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx\) \( = \int {\frac{{{{\left( {{{\sin }^4}x} \right)}^2} - {{\left( {{{\cos }^4}x} \right)}^2}}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} .dx\)…
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