---
title: "Integral of (sin x+cos x)/(9+16 sin 2x)"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-sinx-cosx-over-9-16-sin2x"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-sinx-cosx-over-9-16-sin2x
site: BCT Engineering
---

# Integral of (sin x+cos x)/(9+16 sin 2x)

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2072 Kartik
- Video: [Integrate  (sinx+cosx)/(9+16 sin2x)  dx](https://www.youtube.com/watch?v=_IbNOMWS-og) — Maths Wallah- Ratnesh
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-sinx-cosx-over-9-16-sin2x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Evaluate ∫₀^{π/4} (sin x + cos x)/(9 + 16 sin 2x) dx.

## Short answer

Substitute t = sin x – cos x, dt = (cos x + sin x)dx, sin 2x = 1 – t². Limits become -1 to 0. Integral = ∫_{-1}^0 dt/(25 – 16t²) = (1/20)ln 3.

## Answer

Let I = ∫₀^(π/4) (sin x + cos x)/(9 + 16 sin 2x) dx.

Put t = sin x − cos x. Then dt = (cos x + sin x) dx, and

t² = 1 − 2 sin x cos x = 1 − sin 2x, so sin 2x = 1 − t².

Limits: x = 0 → t = −1; x = π/4 → t = 0.

```text
I = ∫_(−1)^0 dt / (9 + 16(1 − t²))
  = ∫_(−1)^0 dt / (25 − 16t²)
  = ∫_(−1)^0 dt / (5² − (4t)²)
```
Use ∫ dt/(c² − k²t²) = (1/(2ck)) ln((c + kt)/(c − kt)) with c = 5, k = 4:

```text
I = (1/40) [ln((5 + 4t)/(5 − 4t))]_(−1)^0
  = (1/40) [ln 1 − ln(1/9)]
  = (1/40) ln 9
  = (1/20) ln 3
```
**Answer: I = (1/20) ln 3 ≈ 0.0549.**

## Check yourself

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

1. The substitution is
   A. t = sin x + cos x
   B. t = sin x - cos x
   C. t = sin 2x
   D. t = tan x
   Answer: B. dt = (cos x + sin x) dx matches the numerator. (Key point: Substitution)

2. sin 2x in terms of t is
   A. t^2 - 1
   B. 1 + t^2
   C. 1 - t^2
   D. 2t
   Answer: C. t^2 = 1 - sin 2x. (Key point: Identity)

3. The limits x = 0 to π/4 become
   A. t = 0 to 1
   B. t = 1 to 0
   C. t = -1 to 1
   D. t = -1 to 0
   Answer: D. t(0) = -1, t(π/4) = 0. (Key point: Limits)

4. The value is
   A. (1/20) ln 3
   B. (1/40) ln 3
   C. (1/20) ln 9
   D. (1/5) ln 3
   Answer: A. (1/40) ln 9 = (1/20) ln 3. (Key point: Result)

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