Question

Assertion (A): \(\displaystyle \int_0^{\frac{\pi}{2}} (\sin^6 x + \cos^6 x)\, dx\) 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 \(\dfrac{\pi}{2}\)

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is not the correct explanation of A
• C. A is true, R is false
• D. A is false, R is true

Hint:

To simplify powers of trigonometric functions, express them in terms of double angles or known identities before integration.

Answer 1

The Correct Option is B

We know: \[ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 \] Using identity: \[ a^3 + b^3 = (a + b)^3 - 3ab(a + b) \] Since \(\sin^2 x + \cos^2 x = 1\), we get: \[ \sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x = 1 - \dfrac{3}{4} \sin^2(2x) \] Now integrate: \[ \int_0^{\frac{\pi}{2}} \left(1 - \dfrac{3}{4} \sin^2(2x)\right) dx \] Using: \[ \int \sin^2(ax) dx = \frac{x}{2} - \frac{\sin(2ax)}{4a} \] We get a value between \(\frac{\pi}{8}\) and \(\frac{\pi}{2}\), so Assertion is correct. Also, \(\sin^6 x + \cos^6 x\) is periodic, but with period \(\pi\), not \(\frac{\pi}{2}\), so R is true but not the correct reason.