Question

The value of \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2x}{1+\pi^{\sin x}} dx\) is :

• A. \(\frac{3\pi}{2}\)
• B. \(\frac{3\pi}{4}\)
• C. \(\frac{\pi}{2}\)
• D. \(\frac{5\pi}{4}\)

Hint:

Whenever you see a definite integral of the form \(\int_{a}^{b} \frac{f(x)}{1+c^{g(x)}} dx\) where \(g(a+b-x) = -g(x)\) and \(f(a+b-x) = f(x)\), the "King's property" is the go-to method. Adding the original integral and the transformed one often leads to a simple integrand.

Answer 1

The Correct Option is B

Step 1: Understanding the Question
We need to evaluate a definite integral over a symmetric interval \([-a, a]\). The integrand has a term of the form \(a^{f(x)}\) in the denominator, which suggests using the "King's property" of definite integrals.
Step 2: Key Formula or Approach
We use the property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx\). For an interval \([-a, a]\), this becomes \(\int_{-a}^{a} f(x) dx = \int_{-a}^{a} f(-x) dx\).
Step 3: Detailed Explanation
Let the given integral be \(I\). \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2x}{1+\pi^{\sin x}} dx \quad \cdots (1) \] Apply the property \( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx \). Here \(a = -\frac{\pi}{2}, b = \frac{\pi}{2}\), so \(a+b=0\). We replace \(x\) with \(-x\). \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2(-x)}{1+\pi^{\sin(-x)}} dx \] Since \(\sin(-x) = -\sin x\) and \(\sin^2(-x) = (-\sin x)^2 = \sin^2 x\), we get: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2x}{1+\pi^{-\sin x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2x}{1+\frac{1}{\pi^{\sin x}}} dx \] \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1+\sin^2x)\pi^{\sin x}}{\pi^{\sin x}+1} dx \quad \cdots (2) \] Now, we add equations (1) and (2): \[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{1+\sin^2x}{1+\pi^{\sin x}} + \frac{(1+\sin^2x)\pi^{\sin x}}{1+\pi^{\sin x}} \right) dx \] \[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1+\sin^2x)(1+\pi^{\sin x})}{1+\pi^{\sin x}} dx \] \[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1+\sin^2x) dx \] Using the identity \(\sin^2x = \frac{1-\cos(2x)}{2}\): \[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(1 + \frac{1-\cos(2x)}{2}\right) dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{3}{2} - \frac{\cos(2x)}{2}\right) dx \] \[ 2I = \left[ \frac{3}{2}x - \frac{\sin(2x)}{4} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \] \[ 2I = \left( \frac{3}{2}\left(\frac{\pi}{2}\right) - \frac{\sin(\pi)}{4} \right) - \left( \frac{3}{2}\left(-\frac{\pi}{2}\right) - \frac{\sin(-\pi)}{4} \right) \] Since \(\sin(\pi) = 0\) and \(\sin(-\pi) = 0\): \[ 2I = \left( \frac{3\pi}{4} - 0 \right) - \left( -\frac{3\pi}{4} - 0 \right) = \frac{3\pi}{4} + \frac{3\pi}{4} = \frac{6\pi}{4} = \frac{3\pi}{2} \] \[ I = \frac{3\pi}{4} \] Step 4: Final Answer
The value of the integral is \(\frac{3\pi}{4}\).