(a) Evaluate ∫₀π/4 log(1 + tan x) dx. OR (b) Find ∫ fra — CBSE Class 12 Mathematics 2023 PYQ

Question

\int_0^{\pi/4} \log(1 + \tan x)\, dx

(OR)

(b) Find

\int \frac{dx}{\sqrt{\sin^3 x \, \cos(x - \alpha)}}

CBSECBSE Class XII Board 2023Subjective· 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 — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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$\int_0^{\pi/4}\log(1+\tan x)\,dx=\tfrac\pi8\log2$ .
$\int\frac{dx}{\sqrt{\sin^3x\cos(x-a)}}=-2\sec a\sqrt{\tfrac{\cos(x-a)}{\sin x}}+C.$

Part (a): $\int_0^{\pi/4}\log(1+\tan x)\,dx$

1. Apply the King property $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ with $a=\tfrac\pi4$ :

I=\int_0^{\pi/4}\log\!\Big(1+\tan\big(\tfrac\pi4-x\big)\Big)dx.

2. Simplify. Since $\tan(\tfrac\pi4-x)=\dfrac{1-\tan x}{1+\tan x}$ ,

1+\tan\!\big(\tfrac\pi4-x\big)=\frac{(1+\tan x)+(1-\tan x)}{1+\tan x}=\frac{2}{1+\tan x},

so $\log(\cdots)=\log 2-\log(1+\tan x)$ .

3. Form an equation.

$I=\int_0^{\pi/4}\big[\log2-\log(1+\tan x)\big]dx=\tfrac\pi4\log2-I.$ …

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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 58 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\int \frac{3ax}{b^2 + c^2x^2} dx = A \log |b^2 + c^2x^2| + K$
›Reveal solutionSolution
The integral fits the pattern $\int \frac{du}{u} = \log|u| + C$ after a substitution. The constant $A$ turns out to be $\frac{3a}{2c^2}$ , which corresponds to option (D).

The problem gives you the result of an integral and asks you to identify the constant $A$ that makes the equation true. This is a classic "match the form" question — you don't need to guess; you just need to perform the integration carefully and compare.

The key insight is that the integrand $\frac{3ax}{b^2 + c^2x^2}$ is a rational function where the numerator is almost the derivative of the denominator. The derivative of $b^2 + c^2x^2$ is $2c^2x$ . Our numerator is $3ax$ , which is a constant multiple of $x$ . So a simple substitution $u = b^2 + c^2x^2$ will turn the integral into $\int \frac{du}{u}$ .

Let's work through it step by step.

Set up the substitution.

Let $u = b^2 + c^2x^2$ . Then $du = 2c^2x \, dx$ , so $x \, dx = \frac{du}{2c^2}$ .

Rewrite the integral in terms of $u$ .

The integral is $\int \frac{3ax}{b^2 + c^2x^2} \, dx = \int \frac{3a}{u} \cdot (x \, dx)$ .

Substitute $x \, dx = \frac{du}{2c^2}$ :

$\int \frac{3a}{u} \cdot \frac{du}{2c^2} = \frac{3a}{2c^2} \int \frac{du}{u}.$

Integrate. $\int \frac{du}{u} = \log|u| + C$ , so

$\frac{3a}{2c^2} \log|u| + C = \frac{3a}{2c^2} \log|b^2 + c^2x^2| + K,$
where $K$ is the constant of integration (we renamed $C$ to $K$ to match the problem).

Compare with the given form. …
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CBSE 2026Set CX1 mark
$\int x^{2}\tan(x^{3}+2)\,dx$
›Reveal solutionSolution

Substitute $u=x^{3}+2$ ; the integral becomes $\tfrac13\int\tan u\,du=\tfrac13\ln|\sec(x^{3}+2)|+C$ .

Concept: The factor $x^{2}$ is (up to a constant) the derivative of the inner function $x^{3}+2$ , so substitution works.

Let $u=x^{3}+2\Rightarrow du=3x^{2}\,dx\Rightarrow x^{2}\,dx=\dfrac{du}{3}$ .

…

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CBSE 2026Set A1 markMCQ
$\int\frac{\tan(\sin^{-1}x)}{\sqrt{1-x^2}}\,dx =$
(a) $\log|\sec(\sin^{-1}x)| + k$
(b) $\log|\cos(\sin^{-1}x)| + k$
(c) $\tan(\sin^{-1}x) + k$
(d) $\log|\sin^{-1}x| + k$
›Reveal solutionSolution

With $u=\sin^{-1}x$ (so $du=\tfrac{dx}{\sqrt{1-x^2}}$ ) the integral is $\int\tan u\,du=\log|\sec u|+k$ .

Let $u=\sin^{-1}x$ . Then $du=\dfrac{dx}{\sqrt{1-x^2}}$ , so

…

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CBSE 2026Set A1 markMCQ
$\int\frac{dx}{e^x + e^{-x}} =$
(a) $\cot^{-1}(e^x) + k$
(b) $\tan^{-1}(e^x) + k$
(c) $\log|e^x + 1| + k$
(d) $\sin^{-1}(e^x) + k$
›Reveal solutionSolution

Substitute $t=e^x$ : $\int\tfrac{dx}{e^x+e^{-x}}=\int\tfrac{dt}{1+t^2}=\tan^{-1}(e^x)+k$ .

Multiply numerator and denominator by $e^x$ :

$\dfrac{1}{e^x+e^{-x}}=\dfrac{e^x}{e^{2x}+1}$ .

Let $t=e^x$ , $dt=e^x\,dx$ . Then

…

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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 ANNUAL1 markMCQ
$\int \sin(2x+3)\,dx =$
(a) $\cos(2x+3)+C$
(b) $-\dfrac{\cos(2x+3)}{2}+C$
(c) $\tan 2x + C$
(d) None of these
›Reveal solutionSolution

$\int \sin(ax+b)\,dx = -\dfrac{\cos(ax+b)}{a}+C$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\int_1^{e} \dfrac{(\log x)^2}{x}\, dx =$
(a) $\dfrac{1}{3}e^3$
(b) $\dfrac{1}{3}(e^3-1)$
(c) $\dfrac{1}{3}$
(d) None of these
›Reveal solutionSolution

Substitute $u = \log x$ , $du = dx/x$ , converting the limits from $x=1,e$ to $u=0,1$ .

Let $u=\log x \Rightarrow du = \dfrac{dx}{x}$ . When $x=1, u=0$ ; when $x=e, u=1$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{1}{x(1+\log x)}\,dx$
(a) $x + \log x + c$
(b) $|x + \log x| + c$
(c) $\log|1 + \log x| + c$
(d) $\log(1+x) + c$
›Reveal solutionSolution

Substitute $u=1+\log x$ so $du=\dfrac{dx}{x}$ , turning the integral into $\int\dfrac{du}{u}$ .

$I=\displaystyle\int\dfrac{1}{x(1+\log x)}\,dx$

Let $u=1+\log x \Rightarrow du=\dfrac{1}{x}dx$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{\sin^6 x}{\cos^8 x}\,dx$
›Reveal solutionSolution

Rewrite the integrand as $\tan^6x\sec^2x$ and substitute $t=\tan x$ .

$I=\displaystyle\int\dfrac{\sin^6x}{\cos^8x}\,dx=\int\dfrac{\sin^6x}{\cos^6x}\cdot\dfrac{1}{\cos^2x}\,dx=\int\tan^6x\sec^2x\,dx$

Let $t=\tan x \Rightarrow dt=\sec^2x\,dx$ . …

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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 2026Set ANNUAL1 markMCQ
$Q.\int x^2 e^{x^3} dx equals:$
(a)
(i) \frac{e^{x^3}}{3} + c
(b)
(ii) 3e^{x^3} + c
(c)
(iii) \frac{e^{x^2}}{3} + c
(d)
(iv) \frac{1}{2}e^{x^2} + c
›Reveal solutionSolution
$\displaystyle\int x^2 e^{x^3}\,dx=\dfrac{e^{x^3}}{3}+c$ — option (i).

Concept. Substitution ( $u$ -substitution): choose $u$ whose derivative already appears (up to a constant) in the integrand.

Steps.

Let $u=x^3$ , then $du=3x^2\,dx$ , i.e. $x^2\,dx=\dfrac{1}{3}\,du$ . …
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Q.(a) Evaluate ∫0π/4log⁡(1+tan⁡x) dx\int_0^{\pi/4} \log(1 + \tan x)\, dx∫0π/4​log(1+tanx)dx.

Part (a): ∫0π/4log⁡(1+tan⁡x) dx\int_0^{\pi/4}\log(1+\tan x)\,dx∫0π/4​log(1+tanx)dx

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Q.(a) Evaluate $\int_0^{\pi/4} \log(1 + \tan x)\, dx$ .

Part (a): $\int_0^{\pi/4}\log(1+\tan x)\,dx$

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