---
title: "Integral of cos x/((1+sin x)(2+sin x))"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-cosx-over-sin-factors"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-cosx-over-sin-factors
site: BCT Engineering
---

# Integral of cos x/((1+sin x)(2+sin x))

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2074 Asoj
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-cosx-over-sin-factors
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Integrate: ∫₀^{π/2} cos x dx / ((1+sin x)(2+sin x)).

## Short answer

Let t = sin x, then integral becomes ∫₀¹ dt/((1+t)(2+t)). Partial fractions give ln((1+t)/(2+t)) evaluated from 0 to 1, yielding ln(4/3).

## Answer

Let I = ∫₀^(π/2) cos x dx / ((1 + sin x)(2 + sin x)).

Put t = sin x, dt = cos x dx. Limits: x = 0 → t = 0; x = π/2 → t = 1.

I = ∫₀¹ dt / ((1 + t)(2 + t))

**Partial fractions:** 1/((1 + t)(2 + t)) = 1/(1 + t) − 1/(2 + t).

```text
I = [ln(1+t) − ln(2+t)]₀¹
  = [ln((1+t)/(2+t))]₀¹
  = ln(2/3) − ln(1/2)
  = ln(4/3)
```
**Answer: I = ln(4/3) ≈ 0.2877.**

## Check yourself

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

1. Put t = sin x; the integral becomes
   A. integral from 0 to π/2 of dt/(1+t)
   B. integral from 0 to 1 of t dt/((1+t)(2+t))
   C. integral from 0 to 1 of dt/((1+t)(2+t))
   D. integral from 1 to 2 of dt/((1+t)(2+t))
   Answer: C. cos x dx = dt, with limits 0 and 1. (Key point: Substitution)

2. 1/((1+t)(2+t)) equals
   A. 1/(1+t) + 1/(2+t)
   B. 1/(2+t) - 1/(1+t)
   C. 2/(1+t) - 1/(2+t)
   D. 1/(1+t) - 1/(2+t)
   Answer: D. Check: (2+t) - (1+t) = 1. (Key point: Partial fractions)

3. The value is
   A. ln(4/3)
   B. ln(3/2)
   C. ln(2/3)
   D. ln 2
   Answer: A. ln(2/3) - ln(1/2) = ln(4/3). (Key point: Result)

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