CBSE Sample Paper 2023-24 · Section C · 3 marks

Q28: Find $\int\sqrt{\frac{x}{1 - x^3}}\,dx$ ⁠, or evaluate $\int_0^{\frac{\pi}{4}}\log(1 + \tan x)\,dx$

Sample 2023-24Ch 7 · Integrals

The question

(a) Find : $\displaystyle\int\sqrt{\frac{x}{1 - x^3}}\,dx$ ⁠; $x \in (0, 1)$ ⁠.

(b) Evaluate: $\displaystyle\int_0^{\frac{\pi}{4}}\log(1 + \tan x)\,dx$ ⁠.

Exam answer

(a)

Let $x^{\frac32} = t$ ⁠, so $x^3 = t^2$ and

Why?

Write the integrand as $\frac{\sqrt x}{\sqrt{1 - x^3}}$ ⁠. Two things suggest $t = x^{\frac32}$ ⁠:

$x^3 = \left(x^{\frac32}\right)^2 = t^2$ ⁠, so the bottom becomes $\sqrt{1 - t^2}$ ⁠, a standard form.

The derivative of $x^{\frac32}$ contains $x^{\frac12} = \sqrt x$ ⁠, which is exactly the top.

dt = \frac32x^{\frac12}\,dx \Rightarrow \sqrt x\,dx = \frac23\,dt

Why?

Power rule: $\frac{d}{dx}x^{\frac32} = \frac32x^{\frac32 - 1} = \frac32x^{\frac12}$ ⁠.

Multiply both sides by $\frac23$ to get $\sqrt x\,dx$ on its own.

\begin{aligned} dt &= \frac32x^{\frac12}\,dx \\ \frac23\,dt &= \frac23 \cdot \frac32x^{\frac12}\,dx \\ \frac23\,dt &= x^{\frac12}\,dx \quad \left(\frac23 \cdot \frac32 = 1\right) \\ \frac23\,dt &= \sqrt x\,dx \end{aligned}

\begin{aligned} \int\sqrt{\frac{x}{1 - x^3}}\,dx &= \int\frac{\sqrt x\,dx}{\sqrt{1 - x^3}} \\ &= \frac23\int\frac{dt}{\sqrt{1 - t^2}} \\ &= \frac23\sin^{-1}(t) + c \end{aligned}

Why?

Since $x \in (0, 1)$ ⁠, both $x$ and $1 - x^3$ are positive, so the root of the fraction splits into a root on top and a root below.

Now replace $\sqrt x\,dx$ by $\frac23\,dt$ and $x^3$ by $t^2$ ⁠:

\begin{aligned} \int\frac{\sqrt x\,dx}{\sqrt{1 - x^3}} &= \int\frac{\frac23\,dt}{\sqrt{1 - t^2}} \\ &= \frac23\int\frac{dt}{\sqrt{1 - t^2}} \end{aligned}

The standard integral used is

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

Finally put $t = x^{\frac32}$ back, so the answer is in $x$ ⁠.

$= \frac23\sin^{-1}\left(x^{\frac32}\right) + c$ ⁠, where $c$ is an arbitrary constant of integration.

(b)

I = \int_0^{\frac{\pi}{4}}\log(1 + \tan x)\,dx\quad\ldots(\text{i})

Why?

$\log(1 + \tan x)$ has no standard antiderivative. The plan is to use a property that brings the same integral $I$ back on the right side, and then solve for $I$ ⁠. That is why it is named and labelled (i).

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

\begin{aligned} I &= \int_0^{\frac{\pi}{4}}\log\left(1 + \tan\left(\frac{\pi}{4} - x\right)\right)dx \\ &= \int_0^{\frac{\pi}{4}}\log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right)dx \end{aligned}

Why?

Here $a = \frac{\pi}{4}$ ⁠, so every $x$ is replaced by $\frac{\pi}{4} - x$ ⁠.

Then use the formula

\begin{aligned} &\tan(A - B) \\ &= \frac{\tan A - \tan B}{1 + \tan A\tan B} \end{aligned}

with $A = \frac{\pi}{4}$ ⁠, $\tan\frac{\pi}{4} = 1$ and $B = x$ ⁠.

\begin{aligned} \tan\left(\frac{\pi}{4} - x\right) &= \frac{1 - \tan x}{1 + (1)\tan x} \\ &= \frac{1 - \tan x}{1 + \tan x} \end{aligned}

\begin{aligned} I &= \int_0^{\frac{\pi}{4}}\log\left(\frac{2}{1 + \tan x}\right)dx \\ &= \int_0^{\frac{\pi}{4}}\log 2\,dx \\ &\quad - \int_0^{\frac{\pi}{4}}\log(1 + \tan x)\,dx \\ &= \frac{\pi}{4}\log 2 - I \quad[\text{using (i)}] \end{aligned}

Why?

Add the fractions inside the log over the common denominator $1 + \tan x$ ⁠:

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

Then $\log\frac{m}{n} = \log m - \log n$ splits the integral in two.

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

$\log 2$ is a constant, so its integral is $\log 2 \times \left(\frac{\pi}{4} - 0\right)$ ⁠. The second integral is $I$ itself, by (i).

\Rightarrow 2I = \frac{\pi}{4}\log 2

Why?

Move $-I$ to the left: $I + I = 2I$ ⁠. Then divide by 2.

\begin{aligned} I &= \frac{\pi}{4}\log 2 - I \\ I + I &= \frac{\pi}{4}\log 2 \quad (\text{add } I) \\ 2I &= \frac{\pi}{4}\log 2 \\ I &= \frac{\pi}{4 \times 2}\log 2 = \frac{\pi}{8}\log 2 \end{aligned}

Watch out: do not try to integrate $\log(1 + \tan x)$ directly. The property does all the work.

$\therefore\ I = \dfrac{\pi}{8}\log 2$ ⁠.

Where this comes from

Chapter 7, Integrals: Integration by substitution; f(a − x) property.

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: Write the integrand as $\frac{\sqrt x}{\sqrt{1 - x^3}}$ ⁠. Since $x^3 = (x^{3/2})^2$ ⁠, and the derivative of $x^{3/2}$ brings out the $\sqrt x$ on top, put $x^{3/2} = t$ ⁠: the integral becomes $\frac23\int\frac{dt}{\sqrt{1 - t^2}}$ ⁠.

(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 $a = \frac{\pi}{4}$ ⁠, replacing $x$ by $\frac{\pi}{4} - x$ turns $1 + \tan x$ into $\frac{2}{1 + \tan x}$ ⁠. The log then splits into $\log 2$ minus the original integrand, so $I = \frac{\pi}{4}\log 2 - I$ ⁠.

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