Evaluate : $\int_1^3 \frac{\sqrt{4-x}}{\sqrt{x} + \sqrt{4-x}}\, dx$

> [!TLDR] > Use the substitution $x \to 4-x$ to exploit symmetry, adding the two forms of the integral gives a simple constant, leading to the value $\boxed{1}$. The key insight here is that the integrand has a built-in symmetry: swapping $x$ with $4-x$ turns the denominator into its mirror image while the numerator and denominator exchange roles. This is a classic trick for integrals of the form $\int_a^b \frac{f(x)}{f(x)+f(a+b-x)}\,dx$, where the answer is simply half the length of the interval. Let’s see why that works. --- 1. **Define the integral and apply the substitution $x \to 4-x$.** Let $$ I = \int_1^3 \frac{\sqrt{4-x}}{\sqrt{x} + \sqrt{4-x}}\, dx. $$ Now substitute $t = 4 - x$. When $x = 1$, $t = 3$; when $x = 3$, $t = 1$. Also $dx = -dt$, so the limits reverse: $$ I = \int_{3}^{1} \frac{\sqrt{t}}{\sqrt{4-t} + \sqrt{t}}\, (-dt) = \int_{1}^{3} \frac{\sqrt{t}}{\sqrt{4-t} + \sqrt{t}}\, dt. $$ The variable is dummy, so rename $t$ back to $x$: $$ I = \int_1^3 \frac{\sqrt{x}}{\sqrt{4-x} + \sqrt{x}}\, dx. $$ 2. **Add the two expressions for $I$.** We now have two different-looking integrals that are actually equal: $$ I = \int_1^3 \frac{\sqrt{4-x}}{\sqrt{x} + \sqrt{4-x}}\, dx \quad\text{and}\quad I = \int_1^3 \frac{\sqrt{x}}{\sqrt{4-x} + \sqrt{x}}\, dx. $$ Adding them: …

Evaluate : ∫₁³ frac√(4-x)√(x) + √(4-x) dx

Question

\int_1^3 \frac{\sqrt{4-x}}{\sqrt{x} + \sqrt{4-x}}\, dx

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — King Property of Definite Integrals

The King Property of Definite Integrals

Walk a path from $a$ to $b$ measuring something at each step; now walk it backwards from $b$ to $a$ . The King Property says the total is unchanged — provided you also reverse how you measure. It is one of the most useful shortcuts for definite integrals.

\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx

The limits stay $a$ to $b$ ; only the argument changes, $x \to a+b-x$ .

Where it comes from

Substitute $t = a + b - x$ , so $dx = -dt$ ; when $x=a$ , $t=b$ and when $x=b$ , $t=a$ :

\int_{a}^{b} f(x)\,dx = \int_{b}^{a} f(a+b-t)\,(-dt) = \int_{a}^{b} f(a+b-t)\,dt.

Renaming $t$ back to $x$ gives the result. So it is not a trick — just substitution.

Why it helps

Adding the original integral to its "mirror" often collapses the integrand. For instance, with $I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx$ , the property replaces $\sin x$ by $\cos x$ (since $\sin(\tfrac{\pi}{2}-x)=\cos x$ ). Adding the two forms:

2I = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \frac{\pi}{2}, \qquad I = \frac{\pi}{4}.

Tip

Reach for it when the integrand has $\sin x,\cos x,\tan x$ over $[0,\pi/2]$ or $[0,\pi]$ and $f(a+b-x)$ simplifies. If the swapped form is no easier, it will not help.

Watch out

The limits do not change — only the function's argument does. …

Start your 14-day free trial to unlock the full solution →

Use the substitution $x \to 4-x$ to exploit symmetry, adding the two forms of the integral gives a simple constant, leading to the value $\boxed{1}$ .

The key insight here is that the integrand has a built-in symmetry: swapping $x$ with $4-x$ turns the denominator into its mirror image while the numerator and denominator exchange roles. This is a classic trick for integrals of the form $\int_a^b \frac{f(x)}{f(x)+f(a+b-x)}\,dx$ , where the answer is simply half the length of the interval.

Let’s see why that works.

Define the integral and apply the substitution $x \to 4-x$ . Let

I = \int_1^3 \frac{\sqrt{4-x}}{\sqrt{x} + \sqrt{4-x}}\, dx.

Now substitute $t = 4 - x$ . When $x = 1$ , $t = 3$ ; when $x = 3$ , $t = 1$ . Also $dx = -dt$ , so the limits reverse:

I = \int_{3}^{1} \frac{\sqrt{t}}{\sqrt{4-t} + \sqrt{t}}\, (-dt) = \int_{1}^{3} \frac{\sqrt{t}}{\sqrt{4-t} + \sqrt{t}}\, dt.

The variable is dummy, so rename $t$ back to $x$ :

I = \int_1^3 \frac{\sqrt{x}}{\sqrt{4-x} + \sqrt{x}}\, dx.

Add the two expressions for $I$ . We now have two different-looking integrals that are actually equal:

I = \int_1^3 \frac{\sqrt{4-x}}{\sqrt{x} + \sqrt{4-x}}\, dx \quad\text{and}\quad I = \int_1^3 \frac{\sqrt{x}}{\sqrt{4-x} + \sqrt{x}}\, dx.

Adding them: …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

CBSE 2026Set A1 markMCQ
$\int_0^{\pi/2}\frac{\sin x}{\sin x + \cos x}\,dx =$
(a) $\pi$
(b) $\frac{\pi}{2}$
(c) $0$
(d) $\frac{\pi}{4}$
›Reveal solutionSolution

Add the integral to its $x\to\tfrac{\pi}{2}-x$ image to get $2I=\tfrac{\pi}{2}$ , so $I=\tfrac{\pi}{4}$ .

Let $I=\int_0^{\pi/2}\dfrac{\sin x}{\sin x+\cos x}\,dx$ . Replacing $x$ by $\tfrac{\pi}{2}-x$ swaps $\sin$ and $\cos$ :

$I=\int_0^{\pi/2}\dfrac{\cos x}{\cos x+\sin x}\,dx$ .

Adding the two expressions for $I$ :

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$\int_0^a\frac{\sqrt{x}}{\sqrt{a-x} + \sqrt{x}}\,dx =$
(a) $a$
(b) $\frac{a}{2}$
(c) $2a$
(d) $3a$
›Reveal solutionSolution

Use $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ ; the substitution swaps $\sqrt{x}$ and $\sqrt{a-x}$ , giving $I=\tfrac{a}{2}$ .

Let $I=\int_0^a\dfrac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}\,dx$ . Replacing $x$ by $a-x$ :

$I=\int_0^a\dfrac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx$ .

Adding the two forms:

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. \int (from π/6 to π/3) \sqrt(cos x) / (\sqrt(sin x) + \sqrt(cos x)) dx is equal to:$
(a) π/4
(b) π/6
(c) π/12
(d) π/2
›Reveal solutionSolution

Using the property $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$ , the integral equals its own "partner" integral, so twice the integral equals the length of the interval.

Let $I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$

Using the property $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ with $a=\pi/6, b=\pi/3$ , so $a+b = \pi/2$ :

$I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\sin(\pi/2-x)}+\sqrt{\cos(\pi/2-x)}}\,dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx$

Call this second integral $J$ . By relabeling, $I = J$ .

Adding the original definitions of $I$ and $J$ : …

Start your 14-day free trial to see the full solution →
CBSE 2025Set IX1 markMCQ
$\displaystyle\int_{0}^{\pi/2} \dfrac{dx}{1 + \sqrt{\tan x}}$
(a) $0$
(b) $\dfrac{\pi}{2}$
(c) $\dfrac{\pi}{4}$
(d) $\dfrac{\pi}{8}$
›Reveal solutionSolution

By the king-property $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ , $I=\dfrac{\pi}{4}$ ; option (c).

Concept. The property $\displaystyle\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ turns a hard integral into a solvable pair.

Let $I=\displaystyle\int_0^{\pi/2}\frac{dx}{1+\sqrt{\tan x}}$ . Replace $x$ by $\tfrac{\pi}{2}-x$ ; since $\tan\!\left(\tfrac{\pi}{2}-x\right)=\cot x$ , …

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx =$
(a) $\pi$
(b) $\pi/2$
(c) $\pi/4$
(d) $2\pi$
›Reveal solutionSolution

Using $\int_0^{\pi/2}f(x)dx=\int_0^{\pi/2}f(\tfrac{\pi}{2}-x)dx$ , add the two forms: $2I=\frac{\pi}{2}$ , so $I=\frac{\pi}{4}$ .

Let $I=\int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.$ Replacing $x\to \tfrac{\pi}{2}-x$ swaps $\sin$ and $\cos$ :

$I=\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.$

Adding the two expressions for $I$ :

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int_{0}^{\pi/2} \log\tan x\,dx =$
(a) $\pi/4$
(b) $\pi/2$
(c) $0$
(d) $\pi$
›Reveal solutionSolution

Replacing $x\to\tfrac{\pi}{2}-x$ turns $\log\tan x$ into $\log\cot x=-\log\tan x$ , so $I=-I\Rightarrow I=0$ .

Let $I=\int_0^{\pi/2}\log\tan x\,dx.$ Using $\int_0^{a}f(x)dx=\int_0^a f(a-x)dx$ with $a=\tfrac{\pi}{2}$ :

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. \int (from π/6 to π/3) \sqrt(cos x) / (\sqrt(sin x) + \sqrt(cos x)) dx is equal to:$
(a) π/4
(b) π/6
(c) π/12
(d) π/2
›Reveal solutionSolution

Use the property $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$ : adding the original integral to its "flipped" version gives a constant, and by symmetry the two halves are equal.

Let $I = \displaystyle\int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$ .

Here $a=\pi/6,\ b=\pi/3$ , so $a+b = \pi/2$ . Replacing $x$ by $a+b-x = \dfrac{\pi}{2}-x$ , and using $\cos\!\left(\dfrac\pi2-x\right)=\sin x$ , $\sin\!\left(\dfrac\pi2-x\right)=\cos x$ :

$I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set D1 markMCQ
$\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}}\,dx =$
(a) $a$
(b) $\frac{a}{2}$
(c) $2a$
(d) $3a$
›Reveal solutionSolution

Adding the integral to its $x\to a-x$ image gives $2I=a$ , so $I=\frac{a}{2}$ .

Let $I = \displaystyle\int_0^a \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx$ .

Using $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ :

$I = \displaystyle\int_0^a \dfrac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$f(a+b-x) = f(x)$
(a) $0$
(b) $\dfrac{a+b}{2}\displaystyle\int_a^b f(b+x)\,dx$
(c) $\dfrac{b-a}{2}\displaystyle\int_a^b f(x)\,dx$
(d) $\dfrac{a+b}{2}\displaystyle\int_a^b f(x)\,dx$
›Reveal solutionSolution

This is a standard King's-rule property of definite integrals.

Let $I=\displaystyle\int_a^b x f(x)\,dx$ . Using the property $\int_a^b g(x)dx=\int_a^b g(a+b-x)dx$ :

$I=\displaystyle\int_a^b (a+b-x)f(a+b-x)\,dx=\int_a^b (a+b-x)f(x)\,dx$ (since $f(a+b-x)=f(x)$ )

$I=(a+b)\displaystyle\int_a^b f(x)dx-\int_a^b x f(x)dx=(a+b)\int_a^b f(x)dx-I$

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set FF1 markMCQ
$\displaystyle\int_0^{\pi/2} \dfrac{dx}{1+\sqrt{\tan x}}$
(a) 0
(b) $\dfrac{\pi}{2}$
(c) $\dfrac{\pi}{4}$
(d) $\dfrac{\pi}{8}$
›Reveal solutionSolution

By the King's-rule property, $I=\dfrac{\pi}{4}$ — option (c).

Concept. Use $\displaystyle\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ with $a=\dfrac{\pi}{2}$ .

Let $I=\displaystyle\int_0^{\pi/2}\frac{dx}{1+\sqrt{\tan x}}$ . Replacing $x\to \frac{\pi}{2}-x$ turns $\tan x$ into $\cot x$ :

$I=\int_0^{\pi/2}\frac{dx}{1+\sqrt{\cot x}}=\int_0^{\pi/2}\frac{\sqrt{\tan x}\,dx}{\sqrt{\tan x}+1}.$ …

Start your 14-day free trial to see the full solution →
CBSE 2021Set I1 markMCQ
$\int_0^{\pi/2} \log\cot\theta\,d\theta =$
(a) $\dfrac{\pi}{2}\log 2$
(b) $\dfrac{\pi}{4}\log 2$
(c) $2\pi\log 2$
(d) $0$
›Reveal solutionSolution

$\displaystyle\int_0^{\pi/2} \log\cot\theta\,d\theta = 0$ .

Let $I = \displaystyle\int_0^{\pi/2}\log\cot\theta\,d\theta$ . Apply the King property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ with $a=\dfrac{\pi}{2}$ :

$I = \int_0^{\pi/2}\log\cot\!\left(\dfrac{\pi}{2}-\theta\right)d\theta = \int_0^{\pi/2}\log\tan\theta\,d\theta.$

Adding the two forms: …

Start your 14-day free trial to see the full solution →
CBSE 2021Set I1 markMCQ
$\int_0^{\pi/2} \dfrac{\cos\theta}{\cos\theta + \sin\theta}\,d\theta =$
(a) $\pi$
(b) $\dfrac{\pi}{2}$
(c) $\dfrac{\pi}{3}$
(d) $\dfrac{\pi}{4}$
›Reveal solutionSolution

$\displaystyle\int_0^{\pi/2} \dfrac{\cos\theta}{\cos\theta + \sin\theta}\,d\theta = \dfrac{\pi}{4}$ .

Let $I = \displaystyle\int_0^{\pi/2}\dfrac{\cos\theta}{\cos\theta+\sin\theta}\,d\theta$ . Apply the King property $\theta\to\dfrac{\pi}{2}-\theta$ , which swaps $\sin\leftrightarrow\cos$ :

$I = \int_0^{\pi/2}\dfrac{\sin\theta}{\sin\theta+\cos\theta}\,d\theta.$

Add the two forms: …

Start your 14-day free trial to see the full solution →

Q.Evaluate : ∫134−xx+4−x dx\int_1^3 \frac{\sqrt{4-x}}{\sqrt{x} + \sqrt{4-x}}\, dx∫13​x​+4−x​4−x​​dx

🎓Unlock everything free for 14 days

Q.Evaluate : $\int_1^3 \frac{\sqrt{4-x}}{\sqrt{x} + \sqrt{4-x}}\, dx$

Unlock everything free for 14 days