$\int_0^{\pi/6} \sec^2\left(x - \frac{\pi}{6}\right) dx$ is equal to: (a) $\frac{1}{\sqrt{3}}$ (b) $-\frac{1}{\sqrt{3}}$ (c) $\sqrt{3}$ (d) $-\sqrt{3}$

> [!TLDR] > The integral simplifies by noticing the derivative of $\tan$ is $\sec^2$, so the antiderivative is $\tan\left(x - \frac{\pi}{6}\right)$. Evaluating from $0$ to $\frac{\pi}{6}$ gives $\frac{1}{\sqrt{3}}$, which is option **(a)**. The key here is recognising that $\sec^2$ is the derivative of $\tan$. That means the integral is essentially a direct application of the fundamental theorem — no substitution or trickery needed. The only subtlety is handling the shift inside the argument: $x - \frac{\pi}{6}$ just shifts the tangent function horizontally, which doesn't change the fact that its derivative is $\sec^2$ of the same argument. 1. **Identify the antiderivative.** We know $\frac{d}{dx} \tan u = \sec^2 u \cdot \frac{du}{dx}$. Here $u = x - \frac{\pi}{6}$, so $\frac{du}{dx} = 1$. Therefore, $$ \int \sec^2\left(x - \frac{\pi}{6}\right) dx = \tan\left(x - \frac{\pi}{6}\right) + C. $$ 2. **Apply the limits.** The definite integral from $0$ to $\frac{\pi}{6}$ is: $$ \left[ \tan\left(x - \frac{\pi}{6}\right) \right]_{0}^{\pi/6} = \tan\left(\frac{\pi}{6} - \frac{\pi}{6}\right) - \tan\left(0 - \frac{\pi}{6}\right). $$ 3. **Simplify each term.** - At the upper limit: $\tan(0) = 0$. - At the lower limit: $\tan\left(-\frac{\pi}{6}\right)$. Since $\tan$ is an odd function, $\tan(-\theta) = -\tan\theta$, so $$ \tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}. $$ 4. **Compute the difference.** …

∫₀π/6 sec²(x - (π)/(6)) dx is equal to: (a) frac1√(3) ( — CBSE Class 12 Mathematics 2023 PYQ

Question

\int_0^{\pi/6} \sec^2\left(x - \frac{\pi}{6}\right) dx

(a)

\frac{1}{\sqrt{3}}

(b)

-\frac{1}{\sqrt{3}}

(c)

\sqrt{3}

(d)

-\sqrt{3}

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★

🔒 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 →

The integral simplifies by noticing the derivative of $\tan$ is $\sec^2$ , so the antiderivative is $\tan\left(x - \frac{\pi}{6}\right)$ . Evaluating from $0$ to $\frac{\pi}{6}$ gives $\frac{1}{\sqrt{3}}$ , which is option (a).

The key here is recognising that $\sec^2$ is the derivative of $\tan$ . That means the integral is essentially a direct application of the fundamental theorem — no substitution or trickery needed. The only subtlety is handling the shift inside the argument: $x - \frac{\pi}{6}$ just shifts the tangent function horizontally, which doesn't change the fact that its derivative is $\sec^2$ of the same argument.

Identify the antiderivative. We know $\frac{d}{dx} \tan u = \sec^2 u \cdot \frac{du}{dx}$ . Here $u = x - \frac{\pi}{6}$ , so $\frac{du}{dx} = 1$ . Therefore,

\int \sec^2\left(x - \frac{\pi}{6}\right) dx = \tan\left(x - \frac{\pi}{6}\right) + C.

Apply the limits. The definite integral from $0$ to $\frac{\pi}{6}$ is:

\left[ \tan\left(x - \frac{\pi}{6}\right) \right]_{0}^{\pi/6} = \tan\left(\frac{\pi}{6} - \frac{\pi}{6}\right) - \tan\left(0 - \frac{\pi}{6}\right).

Simplify each term.
- At the upper limit: $\tan(0) = 0$ .
- At the lower limit: $\tan\left(-\frac{\pi}{6}\right)$ . Since $\tan$ is an odd function, $\tan(-\theta) = -\tan\theta$ , so

\tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}.

Compute the difference. …

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 27 on this concept.

CBSE 2026Set V11 markMCQ
$\frac{1}{x\sqrt{x^2-1}}$
(a) $\sin^{-1}x+c$
(b) $\cos^{-1}x+c$
(c) $-\operatorname{cosec}^{-1}x+c$
(d) $\cot^{-1}x+c$
›Reveal solutionSolution

The standard antiderivative is $\sec^{-1}x$ , which equals $-\operatorname{cosec}^{-1}x+c$ ; answer (c).

The standard result is
$\int\frac{dx}{x\sqrt{x^2-1}}=\sec^{-1}|x|+c,\qquad x>1.$
Since $\sec^{-1}x+\operatorname{cosec}^{-1}x=\dfrac{\pi}{2}$ (a constant), we have $\sec^{-1}x=\dfrac{\pi}{2}-\operatorname{cosec}^{-1}x$ , so the antiderivative can equally be written …
Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$\displaystyle\int \dfrac{dx}{\sqrt{x^2+a^2}}$
(a) $\log|\sin x| + c$
(b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$
(c) $\csc^{-1}x$
(d) $\log\left|x+\sqrt{x^2-9}\right| + c$
(e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$
(f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$
(g) $\log|\sec x| + c$ . Write the letter of the option that matches Column A (i).
›Reveal solutionSolution

This is the standard NCERT integral formula $\displaystyle\int\dfrac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|+c$ .

This is one of the standard results proved (in the NCERT text) using the substitution $x = a\tan\theta$ : …

Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$\displaystyle\int \sqrt{x^2+a^2}\,dx$
(a) $\log|\sin x| + c$
(b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$
(c) $\csc^{-1}x$
(d) $\log\left|x+\sqrt{x^2-9}\right| + c$
(e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$
(f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$
(g) $\log|\sec x| + c$ . Write the letter of the option that matches Column A (ii).
›Reveal solutionSolution

This is the standard NCERT formula for $\int\sqrt{x^2+a^2}\,dx$ .

This is a standard result (derived in the NCERT text via integration by parts): …

Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$\displaystyle\int \tan x\,dx$
(a) $\log|\sin x| + c$
(b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$
(c) $\csc^{-1}x$
(d) $\log\left|x+\sqrt{x^2-9}\right| + c$
(e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$
(f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$
(g) $\log|\sec x| + c$ . Write the letter of the option that matches Column A (iii).
›Reveal solutionSolution

Write $\tan x = \sin x/\cos x$ and integrate by substitution $u=\cos x$ .

$\int \tan x\,dx = \int \dfrac{\sin x}{\cos x}\,dx$

Let $u = \cos x$ , $du = -\sin x\,dx$ : …

Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$\displaystyle\int \cot x\,dx$
(a) $\log|\sin x| + c$
(b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$
(c) $\csc^{-1}x$
(d) $\log\left|x+\sqrt{x^2-9}\right| + c$
(e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$
(f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$
(g) $\log|\sec x| + c$ . Write the letter of the option that matches Column A (iv).
›Reveal solutionSolution

Write $\cot x = \cos x/\sin x$ and integrate by substitution $u=\sin x$ .

$\int \cot x\,dx = \int \dfrac{\cos x}{\sin x}\,dx$

Let $u = \sin x$ , $du = \cos x\,dx$ : …

Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$\displaystyle\int \sqrt{x^2-9}\,dx$
(a) $\log|\sin x| + c$
(b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$
(c) $\csc^{-1}x$
(d) $\log\left|x+\sqrt{x^2-9}\right| + c$
(e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$
(f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$
(g) $\log|\sec x| + c$ . Write the letter of the option that matches Column A (v).
›Reveal solutionSolution

This is the standard NCERT formula $\int\sqrt{x^2-a^2}\,dx$ with $a=3$ .

The standard result (by parts, same family as $\int\sqrt{x^2+a^2}\,dx$ but with a sign change) is:

$\int \sqrt{x^2-a^2}\,dx = \dfrac{x}{2}\sqrt{x^2-a^2} - \dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right| + c$

With $a^2 = 9$ (i.e. $a=3$ ): …

Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$\displaystyle\int \dfrac{dx}{\sqrt{x^2-9}}$
(a) $\log|\sin x| + c$
(b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$
(c) $\csc^{-1}x$
(d) $\log\left|x+\sqrt{x^2-9}\right| + c$
(e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$
(f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$
(g) $\log|\sec x| + c$ . Write the letter of the option that matches Column A (vi).
›Reveal solutionSolution

This is the standard NCERT formula $\int \dfrac{dx}{\sqrt{x^2-a^2}}$ with $a=3$ .

The standard result (by substitution $x=a\sec\theta$ ) is:

$\int \dfrac{dx}{\sqrt{x^2-a^2}} = \log\left|x+\sqrt{x^2-a^2}\right| + c$

With $a=3$ : …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$(ax+b)^2$
›Reveal solutionSolution

Guess a cube and adjust the constant so its derivative reproduces $(ax+b)^2$ .

We seek $F(x)$ with $F'(x)=(ax+b)^2$ . Trying $F(x)=(ax+b)^3$ :

$\frac{d}{dx}(ax+b)^3=3(ax+b)^2\cdot a=3a(ax+b)^2,$

which is $3a$ times too large. Dividing by $3a$ :

$\frac{d}{dx}\left[\frac{(ax+b)^3}{3a}\right]=(ax+b)^2.$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$\int_0^1 \frac{1}{\sqrt{1-x^2}} \,dx =$
(a) $\frac{\pi}{2}$
(b) $\frac{\pi}{4}$
(c) $\frac{2\pi}{3}$
(d) $\frac{3\pi}{2}$
›Reveal solutionSolution

Integrate 1/√(1-x^2) to get sin^{-1}x, then evaluate between 0 and 1.

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$\displaystyle\int \dfrac{dx}{\sqrt{9-25x^2}}$
›Reveal solutionSolution

Reduce to the standard form $\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a}$ .

Rewrite the denominator:

$\sqrt{9-25x^2} = \sqrt{25\left(\dfrac{9}{25}-x^2\right)} = 5\sqrt{\left(\dfrac{3}{5}\right)^2 - x^2}.$

So

$\int \dfrac{dx}{\sqrt{9-25x^2}} = \dfrac{1}{5}\int \dfrac{dx}{\sqrt{(3/5)^2 - x^2}} = \dfrac{1}{5}\sin^{-1}\!\dfrac{x}{3/5} + C = \dfrac{1}{5}\sin^{-1}\!\dfrac{5x}{3} + C.$

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set 65/1/11 markMCQ
$\int_0^{\pi/6} \sec^2\left(x - \frac{\pi}{6}\right) dx$
(a) $\frac{1}{\sqrt{3}}$
(b) $-\frac{1}{\sqrt{3}}$
(c) $\sqrt{3}$
(d) $-\sqrt{3}$
›Reveal solutionSolution
The integral simplifies by noticing the derivative of $\tan$ is $\sec^2$ , so the antiderivative is $\tan\left(x - \frac{\pi}{6}\right)$ . Evaluating from $0$ to $\frac{\pi}{6}$ gives $\frac{1}{\sqrt{3}}$ , which is option (a).

The key here is recognising that $\sec^2$ is the derivative of $\tan$ . That means the integral is essentially a direct application of the fundamental theorem — no substitution or trickery needed. The only subtlety is handling the shift inside the argument: $x - \frac{\pi}{6}$ just shifts the tangent function horizontally, which doesn't change the fact that its derivative is $\sec^2$ of the same argument.

Identify the antiderivative. We know $\frac{d}{dx} \tan u = \sec^2 u \cdot \frac{du}{dx}$ . Here $u = x - \frac{\pi}{6}$ , so $\frac{du}{dx} = 1$ . Therefore,

$\int \sec^2\left(x - \frac{\pi}{6}\right) dx = \tan\left(x - \frac{\pi}{6}\right) + C.$

Apply the limits. The definite integral from $0$ to $\frac{\pi}{6}$ is:

$\left[ \tan\left(x - \frac{\pi}{6}\right) \right]_{0}^{\pi/6} = \tan\left(\frac{\pi}{6} - \frac{\pi}{6}\right) - \tan\left(0 - \frac{\pi}{6}\right).$

Simplify each term.

At the upper limit: $\tan(0) = 0$ .

At the lower limit: $\tan\left(-\frac{\pi}{6}\right)$ . Since $\tan$ is an odd function, $\tan(-\theta) = -\tan\theta$ , so

$\tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}.$

Compute the difference. …
Start your 14-day free trial to see the full solution →
CBSE 2023Set E1 markMCQ
$\displaystyle\int\sin\dfrac{3x}{4}\,dx =$
(a) $k-\dfrac{3}{4}\cos\dfrac{3x}{4}$
(b) $k+\dfrac{3}{4}\cos\dfrac{3x}{4}$
(c) $k-\dfrac{4}{3}\cos\dfrac{3x}{4}$
(d) $k+\dfrac{4}{3}\cos\dfrac{3x}{4}$
›Reveal solutionSolution

$\displaystyle\int\sin\dfrac{3x}{4}\,dx=-\dfrac{4}{3}\cos\dfrac{3x}{4}+k$ .

Using $\int\sin(ax)\,dx=-\dfrac{1}{a}\cos(ax)$ with $a=\dfrac{3}{4}$ :

…

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

Q. $\int_0^{\pi/6} \sec^2\left(x - \frac{\pi}{6}\right) dx$ is equal to:

Concept understanding — Antiderivative By Inspection

Antiderivatives by Inspection

The idea

Straight recognition

Guess-and-adjust

The one safeguard …

Q.∫0π/6sec⁡2(x−π6)dx\int_0^{\pi/6} \sec^2\left(x - \frac{\pi}{6}\right) dx∫0π/6​sec2(x−6π​)dx is equal to:

🎓Unlock everything free for 14 days

Q. $\int_0^{\pi/6} \sec^2\left(x - \frac{\pi}{6}\right) dx$ is equal to:

Unlock everything free for 14 days