CBSE Sample Paper 2022-23 · Section C · 3 marks

Q28: $\int_{\pi/6}^{\pi/3}\frac{dx}{1 + \sqrt{\tan x}}$ ⁠, or $\int_0^4|x - 1|\,dx$

Sample 2022-23Ch 7 · Integrals

The question

(a) Evaluate: $\displaystyle\int_{\pi/6}^{\pi/3}\frac{dx}{1 + \sqrt{\tan x}}$

(b) Evaluate: $\displaystyle\int_0^4|x - 1|\,dx$

Exam answer

(a)

\begin{aligned} I &= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1 + \sqrt{\tan x}} \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \\ &\qquad\ldots(\text{i}) \end{aligned}

Why?

Write the tangent in terms of sine and cosine:

\sqrt{\tan x} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}}

then multiply the numerator and the denominator by $\sqrt{\cos x}$ ⁠. Both are positive on $\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$ ⁠.

\begin{aligned} &\frac{1}{1 + \frac{\sqrt{\sin x}}{\sqrt{\cos x}}} \\ &= \frac{1 \times \sqrt{\cos x}}{\left(1 + \frac{\sqrt{\sin x}}{\sqrt{\cos x}}\right)\sqrt{\cos x}} \\ &= \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \end{aligned}

The order in the bottom does not matter: $\sqrt{\cos x} + \sqrt{\sin x} = \sqrt{\sin x} + \sqrt{\cos x}$ ⁠.

The label (i) is needed because this form is added to another form of $I$ later.

Using $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ ⁠, with $a + b = \frac{\pi}{2}$ ⁠, $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ and $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ ⁠,

\begin{aligned} I &= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx \\ &\qquad\ldots(\text{ii}) \end{aligned}

Why?

The property replaces $x$ by $a + b - x$ and keeps the same limits. Here

\begin{aligned} a + b &= \frac{\pi}{6} + \frac{\pi}{3} \\ &= \frac{\pi}{6} + \frac{2\pi}{6} \\ &= \frac{3\pi}{6} = \frac{\pi}{2} \end{aligned}

so every $x$ becomes $\frac{\pi}{2} - x$ ⁠. This turns cosine into sine and sine into cosine, so $\sqrt{\cos x}$ and $\sqrt{\sin x}$ swap places.

Adding (i) and (ii), we get

\begin{aligned} 2I &= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\bigg(\frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \\ &\qquad +\frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}\bigg)dx \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}dx = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\ &= \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \end{aligned}

Why?

The two integrals have the same limits, so their integrands can be added. The two fractions have the same denominator, and the numerators add up to that denominator, so the sum is 1.

\begin{aligned} &\frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} = 1 \\[4pt] &\frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6} \end{aligned}

Watch out: this gives $2I$ ⁠. Divide by 2 to get $I$ ⁠.

\begin{aligned} 2I &= \frac{\pi}{6} \\ I &= \frac{\pi}{6} \div 2 = \frac{\pi}{12} \end{aligned}

$\therefore\ I = \frac{\pi}{12}$

(b) $|x - 1| = 1 - x$ for $0 \le x \le 1$ ⁠, and $|x - 1| = x - 1$ for $1 \le x \le 4$ ⁠.

Why?

The modulus changes form where the inside is zero: $x - 1 = 0$ at $x = 1$ ⁠. By definition

|u| = \begin{cases} u, & u \ge 0 \\ -u, & u < 0 \end{cases}

For $x < 1$ ⁠, $x - 1$ is negative, so $|x - 1| = -(x - 1) = 1 - x$ ⁠. So split the integral at $x = 1$ ⁠.

\begin{aligned} &\int_0^4|x - 1|\,dx \\ &= \int_0^1(1 - x)\,dx + \int_1^4(x - 1)\,dx \\ &= \left[x - \frac{x^2}{2}\right]_0^1 + \left[\frac{x^2}{2} - x\right]_1^4 \\ &= \left(1 - \frac12\right) + (8 - 4) - \left(\frac12 - 1\right) \\ &= \frac12 + 4 + \frac12 \end{aligned}

Why?

Integrate term by term with $\int x\,dx = \frac{x^2}{2}$ ⁠:

\begin{aligned} \int(1 - x)\,dx &= x - \frac{x^2}{2} \\ \int(x - 1)\,dx &= \frac{x^2}{2} - x \end{aligned}

Upper limit minus lower limit in each bracket:

\begin{aligned} &\left[x - \frac{x^2}{2}\right]_0^1 = \left(1 - \frac12\right) - 0 \\ &\left[\frac{x^2}{2} - x\right]_1^4 = (8 - 4) \\ &\qquad - \left(\frac12 - 1\right) \end{aligned}

At $x = 4$ ⁠, $\frac{x^2}{2} = \frac{16}{2} = 8$ ⁠. Watch the sign: $-\left(\frac12 - 1\right) = +\frac12$ ⁠.

\begin{aligned} 1 - \frac12 &= \frac12 \\ 8 - 4 &= 4 \\ \frac12 - 1 &= -\frac12 \\[4pt] \frac12 + 4 + \frac12 &= \frac12 + \frac12 + 4 \\ &= 1 + 4 = 5 \end{aligned}

$\therefore\ \int_0^4|x - 1|\,dx = 5$

Where this comes from

Chapter 7, Integrals: Definite integrals: properties; modulus integrand.

Question type: (a) Limits from 0 to $a$ ; (b) A modulus inside the integral

How to spot it: click to see

(a) Ch 7, type 12. Limits from 0 to $a$ ⁠: the $f(a - x)$ trick

Spot it: replacing $x$ by $a - x$ swaps the parts of the question, like sin and cos when $a$ is $\frac{\pi}{2}$ ⁠.

Here: First write the integrand as $\frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}$ ⁠. The limits add to $\frac{\pi}{2}$ ⁠, so replacing $x$ by $\frac{\pi}{2} - x$ swaps $\sqrt{\sin x}$ and $\sqrt{\cos x}$ ⁠; adding the two forms gives $2I = \frac{\pi}{3} - \frac{\pi}{6}$ ⁠.

(b) Ch 8, type 6. A modulus inside the integral: split at the corner

Spot it: bars inside the integral, like $|x + 3|$ or $x|x|$ ⁠.

Here: The inside $x - 1$ is 0 at $x = 1$ ⁠, which lies between 0 and 4. Split there: $|x - 1| = 1 - x$ on $[0, 1]$ and $x - 1$ on $[1, 4]$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Integrals in past papers: question 39 of 46 · see them all

Q28: ∫π/6π/3dx1+tan⁡x\int_{\pi/6}^{\pi/3}\frac{dx}{1 + \sqrt{\tan x}}∫π/6π/3​1+tanx​dx​⁠, or ∫04∣x−1∣ dx\int_0^4|x - 1|\,dx∫04​∣x−1∣dx

Q28: $\int_{\pi/6}^{\pi/3}\frac{dx}{1 + \sqrt{\tan x}}$ ⁠, or $\int_0^4|x - 1|\,dx$