Question

The definite integral \[ \int_{\pi/2}^{\pi} \frac{\sin x - x \cos x}{x(x + \sin x)} \, dx \] equals

• A. $\log\left(1 + \frac{2}{\pi}\right)$
• B. $\log\left(1 + \frac{\pi}{2}\right)$
• C. $1 + \frac{2}{\pi}$
• D. $1 + \frac{\pi}{2}$

Hint:

Whenever the integrand contains both trigonometric terms and polynomial terms, look to group them in a way that generates the form $\frac{f'(x)}{f(x)}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires evaluating a definite integral.
We will simplify the integrand using algebraic manipulation to split it into integrable fractions.

Step 2: Key Formula or Approach:
We will use algebraic splitting:
\[ \frac{\sin x - x \cos x}{x(x + \sin x)} = \frac{1}{x} - \frac{1 + \cos x}{x + \sin x} \]
And the basic integration rules:
\[ \int \frac{1}{x} \, dx = \ln|x| \]
\[ \int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| \]

Step 3: Detailed Explanation:

• Let us perform the algebraic splitting of the integrand:
\[ \frac{\sin x - x \cos x}{x(x + \sin x)} = \frac{(x + \sin x) - x(1 + \cos x)}{x(x + \sin x)} \]
\[ = \frac{x + \sin x}{x(x + \sin x)} - \frac{x(1 + \cos x)}{x(x + \sin x)} \]
\[ = \frac{1}{x} - \frac{1 + \cos x}{x + \sin x} \]

• Now we can write the definite integral as:
\[ I = \int_{\pi/2}^{\pi} \left( \frac{1}{x} - \frac{1 + \cos x}{x + \sin x} \right) \, dx \]

• Let us find the antiderivative:
- For the first term: $\int \frac{1}{x} \, dx = \ln(x)$.
- For the second term, let $u = x + \sin x$, then $du = (1 + \cos x) \, dx$. This gives:
\[ \int \frac{1 + \cos x}{x + \sin x} \, dx = \ln(x + \sin x) \]

• Therefore, the antiderivative is:
\[ F(x) = \ln(x) - \ln(x + \sin x) = \ln\left( \frac{x}{x + \sin x} \right) \]

• Evaluating at the limits $\frac{\pi}{2}$ and $\pi$:
- At $x = \pi$:
\[ F(\pi) = \ln\left( \frac{\pi}{\pi + \sin\pi} \right) = \ln\left( \frac{\pi}{\pi} \right) = \ln(1) = 0 \]
- At $x = \frac{\pi}{2}$:
\[ F\left(\frac{\pi}{2}\right) = \ln\left( \frac{\pi/2}{\pi/2 + \sin(\pi/2)} \right) = \ln\left( \frac{\pi/2}{\pi/2 + 1} \right) = \ln\left( \frac{\pi}{\pi + 2} \right) \]

• Calculate the final value of the integral $I$:
\[ I = F(\pi) - F\left(\frac{\pi}{2}\right) = 0 - \ln\left( \frac{\pi}{\pi + 2} \right) \]
\[ I = \ln\left( \frac{\pi + 2}{\pi} \right) = \ln\left( 1 + \frac{2}{\pi} \right) \]



Step 4: Final Answer:
The definite integral equals $\log\left(1 + \frac{2}{\pi}\right)$.