CBSE Sample Paper 2021-22 (Term 2) · Section A · 2 marks

Q1: $\int \frac{\log x}{(1 + \log x)^2}\,dx$ ⁠, or $\int \frac{\sin 2x}{\sqrt{9 - \cos^4x}}\,dx$

Sample 2021-22 T2Ch 7 · Integrals

The question

(a) Find $\displaystyle\int \frac{\log x}{(1 + \log x)^2}\,dx$

(b) Find $\displaystyle\int \frac{\sin 2x}{\sqrt{9 - \cos^4x}}\,dx$

Exam answer

(a)

Let $I = \displaystyle\int \frac{\log x}{(1 + \log x)^2}\,dx$ ⁠.

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

Why?

The numerator $\log x$ is written as $(1 + \log x) - 1$ ⁠. Then the fraction splits into two simpler pieces:

\begin{aligned} &\frac{\log x}{(1 + \log x)^2} \\ &= \frac{1}{1 + \log x} - \frac{1}{(1 + \log x)^2} \end{aligned}

Step by step:

\begin{aligned} &\frac{(1 + \log x) - 1}{(1 + \log x)^2} \\ &= \frac{1 + \log x}{(1 + \log x)^2} - \frac{1}{(1 + \log x)^2} \\ &= \frac{1}{1 + \log x} - \frac{1}{(1 + \log x)^2} \end{aligned}

In the first piece, one factor $(1 + \log x)$ cancels from the top and the bottom.

Only the first piece is integrated. The second one cancels later.

Integrating the first integral by parts, with $\frac{1}{1 + \log x}$ as the first function and 1 as the second,

Why?

Integration by parts:

\begin{aligned} \int u\,v\,dx &= u\int v\,dx \\ &\quad - \int\left(\frac{du}{dx}\int v\,dx\right)dx \end{aligned}

Take $u = \frac{1}{1 + \log x}$ ⁠, which is easy to differentiate, and $v = 1$ ⁠, whose integral is simply $x$ ⁠.

\begin{aligned} I &= \frac{1}{1 + \log x}\times x \\ &\quad - \int \frac{-1}{(1 + \log x)^2}\times\frac1x\times x\,dx \\ &\quad - \int \frac{dx}{(1 + \log x)^2} \\ &= \frac{x}{1 + \log x} + \int \frac{dx}{(1 + \log x)^2} \\ &\quad - \int \frac{dx}{(1 + \log x)^2} \end{aligned}

Why?

By the chain rule,

\begin{aligned} &\frac{d}{dx}\left(\frac{1}{1 + \log x}\right) \\ &= \frac{-1}{(1 + \log x)^2}\times\frac1x \end{aligned}

In detail, write $u = (1 + \log x)^{-1}$ ⁠. The outside part is $(\ )^{-1}$ and the inside part is $1 + \log x$ ⁠, whose derivative is $\frac1x$ ⁠:

\begin{aligned} \frac{du}{dx} &= -1(1 + \log x)^{-2} \times \frac1x \\ &= \frac{-1}{(1 + \log x)^2} \times \frac1x \end{aligned}

Also $\int v\,dx = \int 1\,dx = x$ ⁠. Put these into the formula:

\begin{aligned} &u\int v\,dx = \frac{1}{1 + \log x} \times x \\ &\quad = \frac{x}{1 + \log x} \\[4pt] &-\int \frac{-1}{(1 + \log x)^2} \times \frac1x \times x\,dx \\ &\quad = +\int \frac{dx}{(1 + \log x)^2} \end{aligned}

Then $\frac1x \times x = 1$ ⁠, and the two minus signs make a plus. The new integral is exactly the one being subtracted, so the two cancel.

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

Why?

Check by the quotient rule:

\begin{aligned} &\frac{d}{dx}\left(\frac{x}{1 + \log x}\right) \\ &= \frac{(1 + \log x) - x\cdot\frac1x}{(1 + \log x)^2} \\ &= \frac{\log x}{(1 + \log x)^2} \end{aligned}

This is the integrand, so the answer is right. One constant $C$ is written at the end.

(b)

Let $I = \displaystyle\int \frac{\sin 2x}{\sqrt{9 - \cos^4x}}\,dx$ ⁠.

Put $\cos^2x = t$ ⁠. Then $-2\cos x\sin x\,dx = dt$ ⁠, i.e. $\sin 2x\,dx = -dt$ ⁠.

Why?

Look for a function whose derivative is also in the integral. The derivative of $\cos^2x$ is $-\sin 2x$ ⁠, and $\cos^4x = (\cos^2x)^2 = t^2$ ⁠. So $t = \cos^2x$ removes every $x$ ⁠.

\begin{aligned} \frac{d}{dx}(\cos^2x) &= 2\cos x\,(-\sin x) \\ &= -\sin 2x \end{aligned}

Here $2\sin x\cos x = \sin 2x$ ⁠.

So the substitution gives, step by step:

\begin{aligned} dt &= -2\cos x\sin x\,dx \\ dt &= -\sin 2x\,dx \\ \sin 2x\,dx &= -dt \end{aligned}

\begin{aligned} I &= -\int \frac{dt}{\sqrt{3^2 - t^2}} \\ &= -\sin^{-1}\frac{t}{3} + C \end{aligned}

Why?

Under the root, $9 - \cos^4x = 3^2 - t^2$ ⁠. Use the standard integral with $a = 3$ ⁠:

\begin{aligned} 9 - \cos^4x &= 9 - (\cos^2x)^2 \\ &= 9 - t^2 = 3^2 - t^2 \end{aligned}

\int \frac{dt}{\sqrt{a^2 - t^2}} = \sin^{-1}\frac{t}{a} + C

The minus sign comes from $\sin 2x\,dx = -dt$ ⁠. Finally put back $t = \cos^2x$ ⁠.

-\sin^{-1}\frac{t}{3} = -\sin^{-1}\left(\frac{\cos^2x}{3}\right)

$\therefore\ I = -\sin^{-1}\left(\dfrac{\cos^2x}{3}\right) + C$

Where this comes from

Chapter 7, Integrals: Integration by parts; substitution.

Question type: (a) Integration by parts; (b) Substitution

How to spot it: click to see

(a) Type 8. Integration by parts: a product of two different kinds

Spot it: a product such as $x\sin x$ ⁠, $xe^x$ or $x\log x$ ⁠. A lone $\log x$ or $\tan^{-1}x$ works too: take the other function as 1.

Here: No card fits $\frac{\log x}{(1 + \log x)^2}$ ⁠, so write the top as $(1 + \log x) - 1$ and split. Integrate $\frac{1}{1 + \log x}$ by parts with 1 as the second function; the integral it leaves cancels the other piece.

(b) 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: The $\sin 2x$ on top is minus the derivative of $\cos^2 x$ ⁠, and $\cos^4 x = (\cos^2 x)^2$ ⁠. So put $\cos^2 x = t$ ⁠, and the $\sin^{-1}$ card with $a = 3$ finishes $-\int\frac{dt}{\sqrt{3^2 - t^2}}$ ⁠.

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Integrals in past papers: question 41 of 46 · see them all

Q1: ∫log⁡x(1+log⁡x)2 dx\int \frac{\log x}{(1 + \log x)^2}\,dx∫(1+logx)2logx​dx⁠, or ∫sin⁡2x9−cos⁡4x dx\int \frac{\sin 2x}{\sqrt{9 - \cos^4x}}\,dx∫9−cos4x​sin2x​dx

Q1: $\int \frac{\log x}{(1 + \log x)^2}\,dx$ ⁠, or $\int \frac{\sin 2x}{\sqrt{9 - \cos^4x}}\,dx$