Question

\begin{aligned} &\text { Integrating by parts, prove that }\\\ &0<\int_{100 \pi}^{200 \pi} \frac{\cos \mathrm{x}}{\mathrm{x}} \mathrm{dx}<\frac{1}{100 \pi} \end{aligned}

Short answer

Answer: The value of the integral $\int_{100 \pi}^{200 \pi} \frac{\cos x}{x} dx$ is positive and lies between 0 and $\frac{1}{100 \pi}$.

Step-by-step solution

Choose u and dv

Let's choose our functions u and dv as follows: \begin{aligned} u =\frac{1}{x} ,\, dv = \cos x \, dx \end{aligned}

Compute du and v

Now we need to compute du and v. Taking the derivative of u and the integral of dv, we get: \begin{aligned} du = -\frac{1}{x^2} dx ,\, v = \int \cos x \, dx = \sin x \end{aligned}

Apply integration by parts formula

Applying the integration by parts formula, we get: \begin{aligned} \int_{100\pi}^{200\pi} \frac{\cos x}{x} dx &= \left[ \frac{1}{x}\sin x \right]_{100 \pi}^{200 \pi} - \int_{100 \pi}^{200 \pi} \left(-\frac{1}{x^2}\sin x\right) dx \\ &= \left( \frac{1}{200 \pi} \sin(200 \pi) - \frac{1}{100 \pi} \sin(100 \pi) \right) + \int_{100 \pi}^{200 \pi} \frac{\sin x}{x^2} dx \end{aligned}

Evaluate the first term of the integration by parts

The first term has sine of multiples of π, so it's zero: \begin{aligned} \frac{1}{200 \pi} \sin(200 \pi) - \frac{1}{100 \pi} \sin(100 \pi) = 0 \end{aligned}

Analyze the remaining integral

We are left with: \begin{aligned} \int_{100\pi}^{200\pi} \frac{\cos x}{x} dx = \int_{100 \pi}^{200 \pi} \frac{\sin x}{x^2} dx \end{aligned} Since 0≤ sin(x)≤1 for all x, and between 100π and 200π sine oscillates between positive and negative values, the integral is positive. To find the upper boundary, let's consider the largest value of the integrand function, which occurs when sin(x) = 1 and x = 100π, hence: \begin{aligned} \frac{\sin x}{x^2} < \frac{1}{(100 \pi)^2} \end{aligned}

Calculate the upper boundary of the integral

Now let's integrate the upper bound of the inequality: \begin{aligned} \int_{100 \pi}^{200 \pi} \frac{1}{(100 \pi)^2} dx &= \frac{1}{(100 \pi)^2} \int_{100 \pi}^{200 \pi} dx \\ &= \frac{1}{(100 \pi)^2} \left[ x \right]_{100 \pi}^{200 \pi} \\ &= \frac{1}{(100 \pi)^2}(200\pi - 100\pi) \\ &= \frac{100 \pi}{(100 \pi)^2} \\ &= \frac{1}{100 \pi} \end{aligned}

Conclusion

We have shown that the value of the integral is positive and smaller than 1/(100π): \begin{aligned} 0<\int_{100 \pi}^{200 \pi} \frac{\cos \mathrm{x}}{\mathrm{x}}\mathrm{dx}<\frac{1}{100 \pi} \end{aligned}