Board 2023 Class 12 Maths Q6 Solution: ∫ from 0 to π/6 of sec²(x − π/6) dx

The question

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

(a) $\frac{1}{\sqrt{3}}$

(b) $-\frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

(d) $-\sqrt{3}$

Exam answer

\begin{aligned} I &= \int_0^{\frac{\pi}{6}} \sec^2\left(x - \frac{\pi}{6}\right)dx \\ &= \left[\tan\left(x - \frac{\pi}{6}\right)\right]_0^{\frac{\pi}{6}} \end{aligned}

Why?

The derivative of $\tan x$ is $\sec^2 x$ ⁠. Shifting $x$ by a constant does not change this, because the inner derivative is 1:

\begin{aligned} &\int \sec^2(x - a)\,dx \\ &= \tan(x - a) + C \end{aligned}

Here $a = \frac{\pi}{6}$ ⁠. A definite integral needs no $C$ ⁠.

\begin{aligned} I &= \tan\left(\frac{\pi}{6} - \frac{\pi}{6}\right) - \tan\left(0 - \frac{\pi}{6}\right) \\ &= \tan 0 - \tan\left(-\frac{\pi}{6}\right) \\ &= 0 + \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \end{aligned}

Why?

Upper limit minus lower limit. Put $x = \frac{\pi}{6}$ ⁠, then $x = 0$ ⁠, into $x - \frac{\pi}{6}$ ⁠:

\begin{aligned} \frac{\pi}{6} - \frac{\pi}{6} &= 0 \\ 0 - \frac{\pi}{6} &= -\frac{\pi}{6} \end{aligned}

$\tan 0 = 0$ ⁠.

$\tan$ is an odd function, so

\begin{aligned} \tan\left(-\frac{\pi}{6}\right) &= -\tan\frac{\pi}{6} \\ &= -\frac{1}{\sqrt{3}} \end{aligned}

(The value $\tan\frac{\pi}{6} = \tan 30^\circ = \frac{1}{\sqrt{3}}$ ⁠.) So

\begin{aligned} I &= 0 - \left(-\frac{1}{\sqrt{3}}\right) \\ &= 0 + \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \end{aligned}

Watch the sign: subtracting $-\frac{1}{\sqrt{3}}$ adds $\frac{1}{\sqrt{3}}$ ⁠. Getting this wrong gives option (b).

Answer: (a) $\dfrac{1}{\sqrt{3}}$

Where this comes from

Chapter 7, Integrals: Definite integral of a standard form.

Question type: Definite integrals

How to spot it: click to see

Type 10. Definite integrals: integrate, then top minus bottom

Spot it: numbers on the integral sign. The answer is a number, with no $+\,C$ ⁠.

Here: The limits $0$ and $\frac{\pi}{6}$ are on the sign. $\sec^2\left(x - \frac{\pi}{6}\right)$ integrates to $\tan\left(x - \frac{\pi}{6}\right)$ ⁠, so the answer is $\tan 0 - \tan\left(-\frac{\pi}{6}\right)$ ⁠.

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Q6: ∫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

Q6: $\int_0^{\pi/6} \sec^2\left(x - \frac{\pi}{6}\right)dx$