\(\pi \int_{-\pi}\left[\\{2 x(1+\sin x)\\} /\left(1+\cos ^{2} x\right)\right] d x\) is equal to....... (a) 0 (b) (c) \(\left(\pi^{2} / 2\right)\) (d) \(\pi^{2}\)

Short Answer

Expert verified
(a) 0

Step by step solution

01

Simplifying the integrand

Let's first rewrite the integral: \(\pi\int_{-\pi}^{\pi}\frac{2x(1+\sin x)}{1+\cos^2 x}dx\) Our goal is to simplify it to make the integration process easier. Notice that \(1+\cos^2x\) can be written as \(\sin^2x + \cos^2x\) (since \(\sin^2x + \cos^2x = 1\)). Now, we can rewrite the integral as: \(\pi\int_{-\pi}^{\pi}\frac{2x(1+\sin x)}{\sin^2 x + \cos^2 x}dx\)
02

Observing the symmetry of the integrand

We will now observe the symmetry of the integrand \(\frac{2x(1+\sin x)}{\sin^2 x + \cos^2 x}\) The given integral is an even function with respect to x multiplied by an odd function of x (\(2x(1+\sin x)\)) and the denominator is always even. Now, the integrand becomes an odd function overall since: \(odd * even = odd\) Since the given function is odd, we may say that: \(\int_{-a}^{a}f(x) dx = 0\), for any odd function Hence, the integral evaluates to \(0\). So, \(\boxed{\text{(a) 0}}\) is the correct answer.

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