Assertion (A): $\int_0^{\frac{\pi}{2}}\left(\sin ^6 x+\cos ^6 x\right) d x$ lies in the interval…
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 $\frac{\pi}{2}$
Both A and R are true and R is the correct explanation of $A$
Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$
$A$ is true, $R$ is false
A is false, R is true
Solution
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 $\frac{1}{4} \leq 1-\frac{3}{4}(\sin 2 x)^2 \leq 1$ for $x \in\left[0, \frac{\pi}{2}\right]$
$\Rightarrow \int_0^{\pi / 2} \frac{1}{4} d x \leq \int_0^{\pi / 2}\left(1-\frac{3}{4}(\sin 2 x)^2\right) d x \leq \int_0^{\pi / 2} 1 d x$
$\Rightarrow \frac{\pi}{8} < \mathrm{I} < \frac{\pi}{2} \Rightarrow$ Assertion is correct
and $f\left(x+\frac{\pi}{2}\right)=\sin ^6\left(x+\frac{\pi}{2}\right)+\cos ^6\left(x+\frac{\pi}{2}\right)$ $=\sin ^6 x+\cos ^6 x$
$\Rightarrow$ Period of $\sin ^6 x+\cos ^6 x$ is $\frac{\pi}{2}$ since reason $R$ is correct.