---
title: "Integral of 1/(1+√tan x)"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-dx-over-1-plus-sqrt-tan"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-dx-over-1-plus-sqrt-tan
site: BCT Engineering
---

# Integral of 1/(1+√tan x)

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2068 Shrawan
- Video: [Q69 | Integral from π/6 to π/3 of 1 / (1 + √tan x) dx | Integrate 1 / (1 + √tan x) from π/6 to π/3](https://www.youtube.com/watch?v=21ncfI8ozdk) — GRAVITY COACHING INSTITUTE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-dx-over-1-plus-sqrt-tan
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Evaluate: ∫₀^{π/2} dx/(1 + √tan x).

## Short answer

Let I = ∫₀^{π/2} dx/(1+√tan x). Replace x by π/2−x to get I = ∫₀^{π/2} √sin x/(√sin x+√cos x) dx. Adding the two forms combines numerator and denominator, giving 2I = ∫₀^{π/2} dx = π/2, so I = π/4.

## Answer

Let I = ∫₀^(π/2) dx/(1 + √tan x).

Write tan x = sin x/cos x, so 1 + √tan x = (√cos x + √sin x)/√cos x:

I = ∫₀^(π/2) √cos x/(√cos x + √sin x) dx

Replace x by π/2 − x (swap sin and cos):

I = ∫₀^(π/2) √sin x/(√sin x + √cos x) dx

Add the two forms:

2I = ∫₀^(π/2) (√cos x + √sin x)/(√sin x + √cos x) dx = ∫₀^(π/2) dx = π/2

**I = π/4.**

## Check yourself

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

1. 1/(1 + sqrt tan x) equals
   A. sqrt sin x/(sqrt cos x + sqrt sin x)
   B. sqrt cos x/(sqrt cos x + sqrt sin x)
   C. 1/(sqrt sin x + sqrt cos x)
   D. cos x/(cos x + sin x)
   Answer: B. Multiply by sqrt cos x. (Key point: Rewrite)

2. Replacing x by π/2 - x gives
   A. sqrt cos x/(sqrt sin x - sqrt cos x)
   B. tan x
   C. sqrt sin x/(sqrt sin x + sqrt cos x)
   D. 1
   Answer: C. Swaps sin and cos. (Key point: Trick)

3. I equals
   A. π/2
   B. π/8
   C. 1
   D. π/4
   Answer: D. 2I = π/2. (Key point: Result)

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