TS EAMCET · Maths · Definite Integration
If \(f(x)=\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x\), then \(\int_0^{\pi / 4} \frac{\sin ^2 2 x}{f(x)} d x=\)
- A 2
- B \(\frac{2}{3}\)
- C \(\frac{-2}{3}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x\) and \(I=\int_0^{\pi / 4} \frac{\sin ^2 2 x}{f(x)} d x\) \(=\int_0^{\pi / 4} \frac{4 \sin ^2 x \cos ^2 x}{\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x} d x\)…
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