CBSE Sample Paper 2020-21 · Part B: Section III · 2 marks

Q23: $\int \frac{dx}{\cos^2 x(1 - \tan x)^2}$ ⁠, or $\int_0^1 x(1 - x)^n\,dx$

Sample 2020-21Ch 7 · Integrals

The question

(a) Find $\displaystyle\int \frac{1}{\cos^2 x(1 - \tan x)^2}\,dx$

(b) Evaluate $\displaystyle\int_0^1 x(1 - x)^n\,dx$

Exam answer

(a)

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

Why?

Since $\frac{1}{\cos^2 x} = \sec^2 x$ ⁠, the integrand contains $\sec^2 x$ ⁠, which is the derivative of $\tan x$ ⁠. That is the sign to substitute for the expression containing $\tan x$ ⁠.

Put $1 - \tan x = t$ ⁠. Then $-\sec^2 x\,dx = dt$ ⁠.

Why?

Differentiate both sides of the substitution:

\frac{d}{dx}(1 - \tan x) = -\sec^2 x

So $\sec^2 x\,dx = -dt$ ⁠. Watch out for this minus sign; it is where most slips happen.

\begin{aligned} I &= \int \frac{-dt}{t^2} = -\int t^{-2}\,dt \\ &= \frac{1}{t} + C \end{aligned}

Why?

Power rule with $n = -2$ ⁠:

\int t^{-2}\,dt = \frac{t^{-1}}{-1} = -\frac{1}{t}

The minus sign in front then makes it $+\frac{1}{t}$ ⁠. Finally put back $t = 1 - \tan x$ ⁠, since the answer must be in $x$ ⁠.

\begin{aligned} -\int t^{-2}\,dt &= -\left(-\frac{1}{t}\right) + C \\ &= \frac{1}{t} + C \\ &= \frac{1}{1 - \tan x} + C \end{aligned}

$\therefore\ I = \frac{1}{1 - \tan x} + C$

(b)

I = \int_0^1 x(1 - x)^n\,dx

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

Why?

Expanding $(1 - x)^n$ for a general $n$ is not possible term by term. The property replaces $x$ by $a - x$ ⁠, here $1 - x$ ⁠, which moves the power $n$ onto a plain $x$ ⁠.

\begin{aligned} I &= \int_0^1 (1 - x)[1 - (1 - x)]^n\,dx \\ &= \int_0^1 (1 - x)x^n\,dx \end{aligned}

Why?

Every $x$ in $x(1 - x)^n$ is replaced by $1 - x$ ⁠. Inside the bracket, $1 - (1 - x) = x$ ⁠.

\begin{aligned} 1 - (1 - x) &= 1 - 1 + x \\ &= x \end{aligned}

\begin{aligned} I &= \int_0^1 (x^n - x^{n+1})\,dx \\ &= \left[\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2}\right]_0^1 \\ &= \left(\frac{1}{n+1} - \frac{1}{n+2}\right) - 0 \end{aligned}

Why?

Multiply out: $(1 - x)x^n = x^n - x^{n+1}$ ⁠. Then integrate each term with the power rule:

\begin{aligned} (1 - x)x^n &= 1 \cdot x^n - x \cdot x^n \\ &= x^n - x^{n+1} \end{aligned}

\int x^m\,dx = \frac{x^{m+1}}{m+1}

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

At $x = 1$ every power of 1 is 1. At $x = 0$ every power is 0, because $n + 1$ and $n + 2$ are positive.

\begin{aligned} \text{At } 1&: \frac{1^{n+1}}{n+1} - \frac{1^{n+2}}{n+2} \\ &= \frac{1}{n+1} - \frac{1}{n+2} \\[4pt] \text{At } 0&: \frac{0}{n+1} - \frac{0}{n+2} = 0 \end{aligned}

I = \frac{(n+2) - (n+1)}{(n+1)(n+2)}

Why?

Take the common denominator $(n+1)(n+2)$ ⁠. The numerator is $(n+2) - (n+1) = 1$ ⁠.

\begin{aligned} &\frac{1}{n+1} - \frac{1}{n+2} \\ &= \frac{n+2}{(n+1)(n+2)} \\ &\quad - \frac{n+1}{(n+1)(n+2)} \\ &= \frac{(n+2) - (n+1)}{(n+1)(n+2)} \\[4pt] &(n+2) - (n+1) \\ &= n + 2 - n - 1 = 1 \end{aligned}

$\therefore\ I = \dfrac{1}{(n+1)(n+2)}$

Where this comes from

Chapter 7, Integrals: Substitution; property of definite integrals.

Question type: (a) Substitution; (b) Limits from 0 to $a$

How to spot it: click to see

(a) Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like $\log x$ and $\frac{1}{x}$ ⁠, or $x^2 + 1$ and $x$ ⁠.

Here: Writing $\frac{1}{\cos^2 x}$ as $\sec^2 x$ shows the derivative of $\tan x$ beside $(1 - \tan x)^2$ ⁠. So put $1 - \tan x = t$ ⁠; then $\sec^2 x\,dx = -dt$ and the integral is $-\int t^{-2}\,dt$ ⁠.

(b) 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: With $a = 1$ ⁠, replacing $x$ by $1 - x$ swaps $x$ and $1 - x$ ⁠, so the power $n$ moves onto a plain $x$ ⁠. The integral becomes $\int_0^1(x^n - x^{n+1})\,dx$ ⁠, which the power card finishes.

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Integrals in past papers: question 45 of 46 · see them all

Q23: ∫dxcos⁡2x(1−tan⁡x)2\int \frac{dx}{\cos^2 x(1 - \tan x)^2}∫cos2x(1−tanx)2dx​⁠, or ∫01x(1−x)n dx\int_0^1 x(1 - x)^n\,dx∫01​x(1−x)ndx

Q23: $\int \frac{dx}{\cos^2 x(1 - \tan x)^2}$ ⁠, or $\int_0^1 x(1 - x)^n\,dx$