---
title: "Show ∫ x sin x/(1+cos²x) dx = π²/4"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-x-sinx-over-1-cos2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-x-sinx-over-1-cos2
site: BCT Engineering
---

# Show ∫ x sin x/(1+cos²x) dx = π²/4

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2072 Chaitra
- Video: [integrate 0 to pi '(xsinx)/(1+cos^2x) dx  || Evaluate `int_(0)^(pi)(xsinx)/((1+cos^(2)x))dx`.](https://www.youtube.com/watch?v=mKlP4fO5Nv8) — concepts & Methods
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-x-sinx-over-1-cos2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Show that ∫₀^π x sin x/(1+cos²x) dx = π²/4.

## Short answer

Using property f(a-x) gives 2I = π∫_0^π sin x/(1+cos²x)dx. Substitute t = cos x, then integrate ∫_{-1}^1 dt/(1+t²) = π/2, so I = π²/4.

## Answer

Let I = ∫₀^π x sin x/(1 + cos²x) dx. Using ∫₀^a f(x)dx = ∫₀^a f(a − x)dx:

I = ∫₀^π (π − x) sin x/(1 + cos²x) dx

Add the two forms:

2I = π ∫₀^π sin x/(1 + cos²x) dx

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

```text
∫₀^π sin x dx/(1 + cos²x) = ∫_(−1)^1 dt/(1 + t²)
                          = [tan⁻¹ t]_(−1)^1
                          = π/4 + π/4 = π/2
```
So 2I = π · π/2, giving **I = π²/4.** Shown.

## Check yourself

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

1. The property used is
   A. integration by parts only
   B. integral f(x) = integral f(a-x)
   C. differentiation under the integral
   D. periodicity of cos
   Answer: B. It leads to 2I in terms of an integral without x. (Key point: Trick)

2. Adding the two forms gives
   A. 2I = integral x sin x dx
   B. 2I = π^2
   C. 2I = π integral sin x/(1+cos^2 x) dx
   D. I = 2π integral ...
   Answer: C. x + (π - x) = π. (Key point: Adding)

3. With t = cos x the limits and integrand become
   A. integral from 0 to π of dt/(1+t^2)
   B. integral from 1 to -1 of dt/(1-t^2)
   C. integral from 0 to 1 of t dt
   D. integral from -1 to 1 of dt/(1+t^2)
   Answer: D. dt = -sin x dx and limits 1 → -1 flip the sign. (Key point: Substitution)

4. I equals
   A. π^2/4
   B. π^2/2
   C. π/4
   D. π^2/8
   Answer: A. 2I = π · π/2. (Key point: Result)

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