---
title: "DUIS: ∫ log((a+b sin x)/(a−b sin x)) dx/sin x"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:duis-log-a-plus-b-sinx"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/duis-log-a-plus-b-sinx
site: BCT Engineering
---

# DUIS: ∫ log((a+b sin x)/(a−b sin x)) dx/sin x

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2072 Chaitra
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/duis-log-a-plus-b-sinx
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Apply differentiation under integral sign to evaluate ∫₀^{π/2} log((a + b sin x)/(a − b sin x)) · dx/sin x.

## Short answer

Differentiate with respect to b: dI/db = ∫₀^{π/2} [1/(a+b sin x)+1/(a−b sin x)] dx = 2a∫₀^{π/2} dx/(a²−b² sin²x) = π/√(a²−b²). Integrate to get I = π arcsin(b/a)+C. Setting b=0 gives C=0, so I = π arcsin(b/a).

## Answer

Let a > |b| > 0 and I(b) = ∫₀^(π/2) log[(a + b sin x)/(a − b sin x)] · dx/sin x. Treat a as constant and differentiate with respect to b:

```text
dI/db = ∫₀^(π/2) [ sin x/(a + b sin x) + sin x/(a − b sin x) ] · dx/sin x
      = ∫₀^(π/2) [ 1/(a + b sin x) + 1/(a − b sin x) ] dx
      = ∫₀^(π/2) 2a dx/(a² − b² sin²x)
```
**Evaluate** ∫₀^(π/2) dx/(a² − b² sin²x). Divide by cos²x and put t = tan x, so sin²x = t²/(1 + t²), dx/cos²x = dt:

```text
∫₀^∞ dt / (a²(1 + t²) − b² t²)
  = ∫₀^∞ dt / (a² + (a² − b²)t²)
  = (1/(a√(a² − b²))) [tan⁻¹(t√(a²−b²)/a)]₀^∞
  = π / (2a√(a² − b²))
```
Therefore

dI/db = 2a · π/(2a√(a² − b²)) = π/√(a² − b²).

Integrate with respect to b:

I(b) = π sin⁻¹(b/a) + C

When b = 0, the integrand is log 1 = 0 (and the limit is 0), so I(0) = 0 and C = 0.

**Answer: I = π sin⁻¹(b/a).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. dI/db reduces to
   A. integral 2b dx/(a^2 - b^2 sin^2 x)
   B. integral dx/(a^2+b^2 sin^2 x)
   C. integral 2a dx/(a^2 - b^2 sin^2 x)
   D. integral 2a sin x dx
   Answer: C. Sum of the two fractions is 2a/(a^2 - b^2 sin^2 x). (Key point: Derivative)

2. To integrate, divide by cos^2 x and put
   A. t = sin x
   B. t = cos x
   C. t = x/2
   D. t = tan x
   Answer: D. This yields an arctan integral. (Key point: Substitution)

3. dI/db equals
   A. π/sqrt(a^2 - b^2)
   B. π/(a^2 - b^2)
   C. 2π/sqrt(a^2+b^2)
   D. π a/sqrt(a^2-b^2)
   Answer: A. 2a · π/(2a sqrt(a^2-b^2)). (Key point: dI/db)

4. I equals
   A. π tan^-1(b/a)
   B. π sin^-1(b/a)
   C. (π/2) sin^-1(b/a)
   D. π log(b/a)
   Answer: B. Integral of 1/sqrt(a^2-b^2) with I(0)=0. (Key point: Result)

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