---
title: "DUIS: ∫ dx/(a²sin²x+b²cos²x)²"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:duis-dx-over-a2sin2-b2cos2-squared"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/duis-dx-over-a2sin2-b2cos2-squared
site: BCT Engineering
---

# DUIS: ∫ dx/(a²sin²x+b²cos²x)²

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2073 Chaitra; 2071 Chaitra (asked 2 times)
- Video: [DIFFERENTIATION UNDER INTEGRAL SIGN | DUIS | S-1 | ENGINEERING FIRST YEAR | SAURABH DAHIVADKAR](https://www.youtube.com/watch?v=JBUZZNcmmNw) — SAURABH DAHIVADKAR
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/duis-dx-over-a2sin2-b2cos2-squared
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Apply the method of differentiation under integral sign to prove that ∫₀^{π/2} dx/(a²sin²x + b²cos²x)² = π(a²+b²)/(4a³b³).

## Short answer

Start with I(A,B)=∫₀^{π/2} dx/(A sin²x+B cos²x) = π/(2√(AB)). Differentiate w.r.t. A and B, add to get ∫ dx/(A sin²x+B cos²x)² = (π/4)(A+B)/(AB)^{3/2}. Substitute A=a², B=b² to obtain π(a²+b²)/(4a³b³).

## Answer

#### Method
Differentiation under the integral sign: if I(λ) = ∫ f(x, λ) dx with constant limits, then dI/dλ = ∫ ∂f/∂λ dx.

#### Step 1: a standard integral with parameters
Let I(A, B) = ∫₀^(π/2) dx / (A sin²x + B cos²x), A, B > 0. Divide by cos²x and put t = tan x:

I = ∫₀^∞ sec²x dx / (A tan²x + B) = ∫₀^∞ dt/(A t² + B) = (1/√(AB)) [tan⁻¹(t√(A/B))]₀^∞ = π / (2√(AB)).

#### Step 2: differentiate with respect to A and B
```text
∂I/∂A = −∫₀^(π/2) sin²x dx/(A sin²x + B cos²x)² = −(π/4) A^(−3/2) B^(−1/2)
∂I/∂B = −∫₀^(π/2) cos²x dx/(A sin²x + B cos²x)² = −(π/4) A^(−1/2) B^(−3/2)
```
Add the two results. Since sin²x + cos²x = 1:

∫₀^(π/2) dx/(A sin²x + B cos²x)² = (π/4)[A^(−3/2)B^(−1/2) + A^(−1/2)B^(−3/2)] = (π/4)(A + B)/(AB)^(3/2).

#### Step 3: put A = a², B = b²

**∫₀^(π/2) dx/(a² sin²x + b² cos²x)² = π(a² + b²)/(4a³b³).** Proved.

## Check yourself

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

1. Step 1 evaluates I(A,B) = integral dx/(A sin^2 x + B cos^2 x) over 0 to π/2 as
   A. π/(2(A+B))
   B. π sqrt(AB)/2
   C. π/(2 sqrt(AB))
   D. π/(4 sqrt(AB))
   Answer: C. Substituting t = tan x gives (1/sqrt(AB)) tan^-1 and the limit π/2. (Key point: Parameter integral)

2. Differentiating I with respect to A gives
   A. plus the integral of sin^2 x/(...)^2
   B. minus the integral of 1/(...)^2
   C. the integral of cos^2 x/(...)
   D. minus the integral of sin^2 x /(A sin^2 x + B cos^2 x)^2
   Answer: D. d/dA of 1/D = -sin^2 x/D^2. (Key point: Differentiation)

3. Adding the A and B derivatives yields the squared integral because
   A. sin^2 x + cos^2 x = 1
   B. sin x cos x = 1/2
   C. A + B = 1
   D. the cross terms cancel
   Answer: A. Numerators sin^2 and cos^2 add to 1. (Key point: Combining)

4. With A = a^2, B = b^2 the result is
   A. π(a^2+b^2)/(4ab)
   B. π(a^2+b^2)/(4a^3 b^3)
   C. π(a+b)/(4a^3b^3)
   D. π/(4a^3b^3)
   Answer: B. (π/4)(A+B)/(AB)^(3/2). (Key point: Result)

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