(a) Evaluate: ∫π/125π/12 fracdx1+√(cot x) OR (b) Evaluate: ∫-π/6π/2 (sin |x| + cos |x|) dx

Question

\int_{\pi/12}^{5\pi/12} \frac{dx}{1+\sqrt{\cot x}}

(OR)

(b) Evaluate:

\int_{-\pi/6}^{\pi/2} (\sin |x| + \cos |x|)\, dx

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

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Part (a)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. …

Part (b)Concept understanding — Definite Integral Piecewise

Integrating a Piecewise Function

A piecewise function follows different rules on different parts of its domain — for example

f(x) = \begin{cases} x, & 0 \le x \le 1 \\ 2 - x, & 1 < x \le 2 \end{cases}

To find a definite integral $\int_a^b f(x)\,dx$ of such a function, you cannot use a single antiderivative across the whole interval, because there is no single formula for $f$ over $[a,b]$ . The key idea is to split the integral at every point where the rule changes and integrate each piece with its own formula.

The additivity property

The tool that makes this legal is the interval-additivity of the definite integral: for any point $c$ between $a$ and $b$ ,

\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx.

So you place the split points exactly where the definition of $f$ switches, and on each sub-interval you substitute the rule that applies there.

The method

Find the break points — the $x$ -values where the piecewise rule changes (and note whether any lie inside $[a,b]$ ).
Split $\int_a^b$ into one integral per sub-interval.
On each piece, replace $f$ by its formula there and integrate normally.
Add the results.

Example. For the $f$ above,

\int_0^2 f(x)\,dx = \int_0^1 x\,dx + \int_1^2 (2-x)\,dx = \left[\tfrac{x^2}{2}\right]_0^1 + \left[2x - \tfrac{x^2}{2}\right]_1^2 = \tfrac12 + \tfrac12 = 1.

Tip

Functions defined with $|x|$ or the greatest-integer function $[x]$ are secretly piecewise. To evaluate $\int_{-2}^{2} |x|\,dx$ , write $|x| = -x$ on $[-2,0]$ and $|x| = x$ on $[0,2]$ , then split at $0$ . …

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King's property collapses the integrand to $1$ , giving $\dfrac{\pi}{6}$ .
Splitting at $0$ using $\sin|x|=-\sin x$ for $x<0$ gives $\dfrac{7-\sqrt3}{2}$ .

Part (a)

Let $I=\displaystyle\int_{\pi/12}^{5\pi/12}\frac{dx}{1+\sqrt{\cot x}}$ . Writing $\cot x=\dfrac{\cos x}{\sin x}$ ,

I=\int_{\pi/12}^{5\pi/12}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.\qquad(1)

The limits sum to $a+b=\tfrac{\pi}{12}+\tfrac{5\pi}{12}=\tfrac{\pi}{2}$ , so apply the King's property $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$ , using $\sin(\tfrac{\pi}{2}-x)=\cos x$ , $\cos(\tfrac{\pi}{2}-x)=\sin x$ :

I=\int_{\pi/12}^{5\pi/12}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.\qquad(2)

Adding (1) and (2):

$2I=\int_{\pi/12}^{5\pi/12}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx=\int_{\pi/12}^{5\pi/12}1\,dx=\frac{5\pi}{12}-\frac{\pi}{12}=\frac{\pi}{3}.$ …

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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 23 on this concept.

CBSE 2026Set A1 markMCQ
$\int_0^{2\pi}|\sin x|\,dx =$
(a) $2$
(b) $4$
(c) $1$
(d) $3$
›Reveal solutionSolution

By symmetry $\int_0^{2\pi}|\sin x|\,dx=2\int_0^{\pi}\sin x\,dx=4$ .

On $[0,\pi]$ , $\sin x\ge 0$ ; on $[\pi,2\pi]$ , $\sin x\le 0$ so $|\sin x|=-\sin x$ . The two humps have equal area, so

…

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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 $I$ :

…

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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:

…

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CBSE 2026Set A1 markMCQ
$\int_{-2}^2|x|\,dx =$
(a) $4$
(b) $3$
(c) $2$
(d) $0$
›Reveal solutionSolution

Even function: $\int_{-2}^2|x|\,dx=2\int_0^2 x\,dx=4$ .

$|x|$ is even, so $\int_{-2}^{2}|x|\,dx=2\int_0^2|x|\,dx=2\int_0^2 x\,dx$ (since $x\ge0$ on $[0,2]$ ).

…

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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 $J$ : …

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CBSE 2025Set 65/1/11 markMCQ
$\int_{-1}^{1} \frac{|x|}{x}\,dx,\ x \neq 0$
›Reveal solutionSolution

$\dfrac{|x|}{x} = -1$ for $x<0$ and $+1$ for $x>0$ , an odd function, so the integral over the symmetric interval $[-1,1]$ is $0$ — option (B).

Solution
$\frac{|x|}{x} = \begin{cases} +1, & x > 0 \ -1, & x < 0 \end{cases}$
Split the integral at $x = 0$ :

$\int_{-1}^{1} \frac{|x|}{x}\,dx = \int_{-1}^{0}(-1)\,dx + \int_{0}^{1}(1)\,dx = (-1)(0-(-1)) + (1)(1-0) = -1 + 1 = 0.$ …
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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 $\tan\!\left(\tfrac{\pi}{2}-x\right)=\cot x$ , …

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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 $I$ :

…

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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 $a=\tfrac{\pi}{2}$ :

…

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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int_{1}^{2} e^{x+[x]}\,dx$
(i) $(e+1)e$
(ii) $(1-e)e$
(iii) $(e-1)e$
(iv) $(e-1)e^{2}$
›Reveal solutionSolution

On $[1,2)$ the greatest-integer function gives $[x]=1$ , so the integral becomes a simple exponential integral.

For $x \in [1,2)$ , the integer part $[x] = 1$ (the single point $x=2$ does not affect the value of a definite integral). So on this interval:

$e^{x+[x]} = e^{x+1}$

…

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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 $\cos\!\left(\dfrac\pi2-x\right)=\sin x$ , $\sin\!\left(\dfrac\pi2-x\right)=\cos x$ :

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

…

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

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

…

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Q.(a) Evaluate: ∫π/125π/12dx1+cot⁡x\int_{\pi/12}^{5\pi/12} \frac{dx}{1+\sqrt{\cot x}}∫π/125π/12​1+cotx​dx​

Part (a)

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Q.(a) Evaluate: $\int_{\pi/12}^{5\pi/12} \frac{dx}{1+\sqrt{\cot x}}$

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