MHT CET · Maths · Indefinite Integration
\(\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}\) (where \(\mathrm{C}\) is constant of integration.)
- A \(-2 \sin (2 x)+C\)
- B \(\frac{1}{2} \cos (2 x)+C\)
- C \(2 \cos (2 x)+C\)
- D \(-\frac{1}{2} \sin (2 x)+C\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{2} \sin (2 x)+C\)
Step-by-step Solution
Detailed explanation
\(\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x\)
\(=\int \frac{\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^2 x-\cos ^2 x\right)\left(1-2 \sin ^2 x \cdot \cos ^2 x\right)}{1-2 \sin ^2 x \cdot \cos ^2 x} d x\)
\(=\int\left(\sin ^2 x-\cos ^2 x\right) d x\)
\(=\int-\cos 2 x d x=-\frac{1}{2} \sin 2 x+c\)
\(=\int \frac{\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^2 x-\cos ^2 x\right)\left(1-2 \sin ^2 x \cdot \cos ^2 x\right)}{1-2 \sin ^2 x \cdot \cos ^2 x} d x\)
\(=\int\left(\sin ^2 x-\cos ^2 x\right) d x\)
\(=\int-\cos 2 x d x=-\frac{1}{2} \sin 2 x+c\)
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