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

The given integral to evaluate is:

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2x}{1+\pi^{\sin x}} \, dx

To solve this integral, we can use the property of definite integrals associated with symmetry:

If f(x) is a continuous function on [-a, a], then:

\int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx

Therefore, we can write:

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2(-x)}{1+\pi^{\sin (-x)}} \, dx

We know that:

• \sin(-x) = -\sin (x)
• \sin^2(-x) = \sin^2(x)

Substituting these into the integral:

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2x}{1+\pi^{-\sin x}} \, dx

Next, we use the properties of exponents, knowing that:

\pi^{-\sin x} = \frac{1}{\pi^{\sin x}}

Therefore, the fraction becomes:

\frac{1+\sin^2x}{1+\frac{1}{\pi^{\sin x}}} = \frac{(1+\sin^2x)\pi^{\sin x}}{\pi^{\sin x} + 1}

We now have two expressions for the integral:

I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2x}{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

Adding these two equations, we get:

2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1+\sin^2x) \, dx

This simplifies the problem because:

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1+\sin^2x) \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2x \, dx

Since the integral of a constant over [-\frac{\pi}{2}, \frac{\pi}{2}] is:

\pi

And using the well-known result:

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2x \, dx = \frac{\pi}{2}

Therefore:

2I = \pi + \frac{\pi}{2} = \frac{3\pi}{2}

Thus:

I = \frac{3\pi}{4}

The correct answer is \frac{3\pi}{4}.