AP EAMCET · Maths · Definite Integration
Assertion (A): \(\int_0^{\frac{\pi}{2}}\left(\sin ^6 x+\cos ^6 x\right) d x\) lies in the interval \(\left(\frac{\pi}{8}, \frac{\pi}{2}\right)\)
Reason (R): \(\sin ^6 x+\cos ^6 x\) is a periodic function with period \(\frac{\pi}{2}\)
- A Both A and R are true and R is the correct explanation of \(A\)
- B Both \(\mathrm{A}\) and \(\mathrm{R}\) are true but \(\mathrm{R}\) is not the correct explanation of \(\mathrm{A}\)
- C \(A\) is true, \(R\) is false
- D A is false, R is true
Answer & Solution
Correct Answer
(B) Both \(\mathrm{A}\) and \(\mathrm{R}\) are true but \(\mathrm{R}\) is not the correct explanation of \(\mathrm{A}\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_0^{\pi / 2}\left(\sin ^6 x+\cos ^6 x\right) d x\) \(\begin{aligned} & I=\int_0^{\pi / 2}\left(\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3\right) d x \\ & =\int_0^{\frac{\pi}{2}}\left(1-\frac{3}{4}(\sin 2 x)^2\right) d x \end{aligned}\) Since…
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