Question

Prove that, as \(n \rightarrow \infty, \int_{0}^{1} \cos n x \tan ^{-1} x d x \rightarrow 0\).

Short answer

Question: Prove that the integral of the product of cos(nx) and arctan(x) from 0 to 1 approaches zero as n approaches infinity. Answer: By substitution and bounding the cosine function term, we derived the inequality \(-1 \leq \int_{0}^{\frac{\pi}{4}} \cos{n(\tan u)} (\sec^2{u}) du \leq 1\). Applying the Squeeze Theorem as n approaches infinity, we found that the integral converges to 0.

Step-by-step solution

Substituting u function with arctan(x)

Let u = arctan(x), then x = tan(u) and dx = (sec^2(u) du). Therefore, the integral becomes: $$ \int_{0}^{\frac{\pi}{4}} \cos{n(\tan u)} (\sec^2{u}) du $$ Notice that the integral is on the range \(0 \leq u \leq \frac{\pi}{4}\).

Bounding the cosine function term

Recall that \(-1 \leq \cos{n(\tan u)} \leq 1\). Using this bound, we can write the inequality: $$ -\int_{0}^{\frac{\pi}{4}} (\sec^2{u}) du \leq \int_{0}^{\frac{\pi}{4}} \cos{n(\tan u)} (\sec^2{u}) du \leq \int_{0}^{\frac{\pi}{4}} (\sec^2{u}) du $$

Evaluating the integrals of sec^2(u) du

Integrate both sides of the inequality with respect to u: $$ -\left[\tan u\right]_0^{\frac{\pi}{4}} \leq \int_{0}^{\frac{\pi}{4}} \cos{n(\tan u)} (\sec^2{u}) du \leq \left[\tan u\right]_0^{\frac{\pi}{4}} $$ $$ -1 \leq \int_{0}^{\frac{\pi}{4}} \cos{n(\tan u)} (\sec^2{u}) du \leq 1 $$

Applying the Squeeze Theorem

From the inequality, we can use the Squeeze Theorem as n approaches infinity. Since the integral is bounded by a constant sequence, the integral will also approach 0 as n approaches infinity: $$ \lim_{n \to \infty} \int_{0}^{1} \cos n x \tan ^{-1} x d x = 0 $$ Therefore, we have proven that the integral converges to 0 as n approaches infinity.