Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.
∫ from 0 to 1 of (dt / (t - sin t))
(Hint: t ≥ sin t for t ≥ 0)

Accepted Answer

First, analyze the behavior of the integrand near the problematic point, which is at the lower limit t = 0, because the denominator $$ t - \sin t $$ approaches zero there. Use the hint $$ t \geq \sin t $$ for $$ t \geq 0  to $$understand that the denominator $$ t - \sin t  is $$non-negative and approaches zero as $$ t 	o 0^+ . $$This suggests the integrand may have a singularity at 0. Apply the Taylor series expansion for $$ \sin t $$ near 0: $$ \sin t = t - \frac{t^3}{6} + O(t^5) . $$Substitute this into the denominator to approximate $$ t - \sin t \approx \frac{t^3}{6} $$ for small $$ t . $$Rewrite the integrand near 0 using this approximation: $$ \frac{1}{t - \sin t} \approx \frac{1}{\frac{t^3}{6}} = \frac{6}{t^3} . $$This helps to compare the given integral to a simpler integral $$ \int_0^1 \frac{1}{t^3} dt . $$Use the Direct Comparison Test or Limit Comparison Test by comparing $$ \frac{1}{t - \sin t}  to$$ $$ \frac{1}{t^3} $$ near 0. Since $$ \int_0^1 \frac{1}{t^3} dt $$ diverges, conclude about the convergence or divergence of the original integral based on this comparison.

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.
∫ from 0 to 1 of (dt / (t - sin t))
(Hint: t ≥ sin t for t ≥ 0)

Verified video answer for a similar problem:

Key Concepts

Improper Integrals and Convergence

Direct Comparison Test

Limit Comparison Test

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.∫ from 0 to 1 of (dt / (t - sin t))(Hint: t ≥ sin t for t ≥ 0)

Verified video answer for a similar problem:

Key Concepts

Improper Integrals and Convergence

Direct Comparison Test

Limit Comparison Test

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.
∫ from 0 to 1 of (dt / (t - sin t))
(Hint: t ≥ sin t for t ≥ 0)