Question

Assertion (A): \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}} + (\cos x)^{\sqrt{2}}} \, dx = \frac{\pi}{12} \]

Reason (R): \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{f(x)}{f(x) + f\left(\frac{\pi}{2}-x\right)} \, dx = \frac{\pi}{12} \]

• A. A is true, R is true and R is the correct explanation of A
• B. A is true, R is 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:

When solving integrals with symmetry, consider using substitution or symmetry properties of trigonometric functions. In this case, the substitution \( x \to \frac{\pi}{2} - x \) simplifies the integral and leads to the desired result.

Answer 1

The Correct Option is A

The problem involves calculating the integral:

\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}} + (\cos x)^{\sqrt{2}}} \, dx \]

and verifying the given assertions and reason. The key to solving this problem lies in exploiting the symmetry of the integrand and the limits of integration.

First, note the function transformation:
\( f(x) = \frac{(\sin x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}} + (\cos x)^{\sqrt{2}}} \)

Using the substitution \(x = \frac{\pi}{2} - t\), we find:

\( \sin(\frac{\pi}{2} - t) = \cos(t) \) and \( \cos(\frac{\pi}{2} - t) = \sin(t) \)

Thus, the expression under integration transforms as:

\[ f\left(\frac{\pi}{2}-x\right) = \frac{(\cos x)^{\sqrt{2}}}{(\cos x)^{\sqrt{2}} + (\sin x)^{\sqrt{2}}} \]

which simplifies to:

\[ f\left(\frac{\pi}{2}-x\right) = \frac{(\cos x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}} + (\cos x)^{\sqrt{2}}} \]

The assertion and the reason state that:

Assertion (A):

\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}} + (\cos x)^{\sqrt{2}}} \, dx = \frac{\pi}{12} \]

Reason (R):

\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{f(x)}{f(x) + f\left(\frac{\pi}{2}-x\right)} \, dx = \frac{\pi}{12} \]

These are equivalent because the given integrals simplify using symmetry and properties of functions involving \(x\) and \(\frac{\pi}{2}-x\).

Substituting \(x = \frac{\pi}{2} - t\) shows that the symmetry \( f(x) + f\left(\frac{\pi}{2}-x\right) = 1 \). Therefore:

\[ \frac{f(x)}{f(x) + f\left(\frac{\pi}{2}-x\right)} = f(x) \]

Thus:

\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(x) \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{f(x)}{1} \, dx = \frac{\pi}{12} \]

Hence, both Assertion (A) and Reason (R) are true, and R correctly explains A. Therefore, the correct option is:

A is true, R is true and R is the correct explanation of A

Answer 2

We are given the assertion \( A \) and the reason \( R \). We need to verify whether the given integrals are true and whether the reason provided is the correct explanation of the assertion. Assertion (A): \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}} + (\cos x)^{\sqrt{2}}} \, dx = \frac{\pi}{12} \] This integral represents a well-known identity in trigonometry, where such integrals evaluate to \( \frac{\pi}{12} \) for specific powers of \( \sqrt{2} \). Hence, the assertion (A) is true. Reason (R): The second integral in the reason is: \[ I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{f(x)}{f(x) + f\left( \frac{\pi}{2}-x \right)} \, dx \] By using symmetry and the fact that \( f(x) = (\sin x)^{\sqrt{2}} \), this integral simplifies to the same form as the one in assertion (A), and it evaluates to \( \frac{\pi}{12} \). Thus, the reason (R) is a correct explanation for assertion (A). Since both the assertion and the reason are true, and the reason correctly explains the assertion, the correct answer is option (1).