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Chapter 12: Problem 1044

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

Short Answer

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(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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Most popular questions from this chapter

\(1 \int_{0} \sqrt{x} \sqrt{(1-\sqrt{x}) d x}\) is equal to...... (a) \((4 / 105)\) (b) \((8 / 105)\) (c) \((16 / 105)\) (d) \((32 / 105)\)

\((1 / e) \int_{1}[(\log t) /(1+t)] d t+e \int_{1}[(\log t) /(1+t)] d t\) is equal to...... (a) \(\mathrm{e}\) (b) \((1 / e)\) (c) 2 (d) \((1 / 2)\)

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\((\pi / 2) \int_{0} \sqrt{(\sec x+1) d x}\) is equal to \(\ldots \ldots\) (a) 0 (b) \((\pi / 4)\) (c) \((\pi / 2)\) (d) \(\pi\)
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