CBSE Board Examination 2022 (Term 2) · Section B · 3 marks

Q9: $\int \frac{dx}{\sqrt x + \sqrt[3]x}$ ⁠, or $\int_0^{\pi/2} \frac{\cos x}{(1 + \sin x)(4 + \sin x)}\,dx$

Board 2022 T2Ch 7 · Integrals

The question

(a) Find : $\displaystyle\int \frac{dx}{\sqrt x + \sqrt[3]x}$ ⁠.

(b) Evaluate : $\displaystyle\int_0^{\pi/2} \frac{\cos x}{(1 + \sin x)\,(4 + \sin x)}\,dx$ ⁠.

Exam answer

(a)

Let $I = \displaystyle\int \frac{dx}{\sqrt x + \sqrt[3]x}$ ⁠.

Put $x = t^6$ ⁠, so that $dx = 6t^5\,dt$ ⁠, $\sqrt x = t^3$ and $\sqrt[3]x = t^2$ ⁠.

Why?

The two roots have orders 2 and 3. Their LCM is 6, so $x = t^6$ turns both roots into whole powers of $t$ ⁠:

\begin{aligned} \sqrt{t^6} &= t^3 \\ \sqrt[3]{t^6} &= t^2 \end{aligned}

Here $x > 0$ ⁠, so we take $t = x^{1/6} > 0$ ⁠. Differentiating $x = t^6$ gives $dx = 6t^5\,dt$ ⁠.

\begin{aligned} I &= \int \frac{6t^5}{t^3 + t^2}\,dt = 6\int \frac{t^3}{t + 1}\,dt \\ &= 6\int \left[(t^2 - t + 1) - \frac{1}{t + 1}\right]dt \end{aligned}

(dividing $t^3$ by $t + 1$ ⁠)

Why?

Take $t^2$ out of the denominator, $t^3 + t^2 = t^2(t + 1)$ ⁠, and cancel it:

\begin{aligned} \frac{6t^5}{t^2(t + 1)} &= \frac{6t^{5 - 2}}{t + 1} \\ &= \frac{6t^3}{t + 1} \end{aligned}

The top has a higher degree than the bottom, so divide first. Using the sum of cubes,

t^3 + 1 = (t + 1)(t^2 - t + 1)

Check it by multiplying out:

\begin{aligned} &(t + 1)(t^2 - t + 1) \\ &= t^3 - t^2 + t + t^2 - t + 1 \\ &= t^3 + 1 \end{aligned}

so $t^3 = (t + 1)(t^2 - t + 1) - 1$ ⁠. Dividing by $t + 1$ gives the bracket.

\begin{aligned} \frac{t^3}{t + 1} &= \frac{(t + 1)(t^2 - t + 1) - 1}{t + 1} \\ &= (t^2 - t + 1) - \frac{1}{t + 1} \end{aligned}

\begin{aligned} &= 6\bigg[\frac{t^3}{3} - \frac{t^2}{2} + t \\ &\qquad - \log|t + 1|\bigg] + C \\ &= 2t^3 - 3t^2 + 6t \\ &\qquad - 6\log|t + 1| + C \end{aligned}

Why?

Integrate term by term with the standard results

\begin{aligned} \int t^n\,dt &= \frac{t^{n + 1}}{n + 1} \\ \int \frac{dt}{t + 1} &= \log|t + 1| \end{aligned}

\begin{aligned} \int t^2\,dt &= \frac{t^3}{3} \\ \int t\,dt &= \frac{t^2}{2} \\ \int 1\,dt &= t \end{aligned}

then multiply every term by 6.

\begin{aligned} 6 \times \frac{t^3}{3} &= 2t^3 \\ 6 \times \frac{t^2}{2} &= 3t^2 \\ 6 \times t &= 6t \\ 6 \times \log|t + 1| &= 6\log|t + 1| \end{aligned}

Putting back $t = x^{1/6}$ ⁠,

Why?

$t^3 = x^{3/6} = \sqrt x$ and $t^2 = x^{2/6} = \sqrt[3]x$ ⁠.

\begin{aligned} 2t^3 &= 2\sqrt x \\ 3t^2 &= 3\sqrt[3]x \\ 6t &= 6x^{1/6} \\ 6\log|t + 1| &= 6\log|x^{1/6} + 1| \end{aligned}

Since $t > 0$ ⁠, $t + 1 > 0$ ⁠, so the modulus sign can be dropped.

$\therefore\ I = 2\sqrt x - 3\sqrt[3]x + 6x^{1/6}$ $- 6\log(x^{1/6} + 1) + C$

(b)

Let $I = \displaystyle\int_0^{\pi/2} \frac{\cos x}{(1 + \sin x)(4 + \sin x)}\,dx$ ⁠.

Let $\sin x = t$ ⁠. Then $\cos x\,dx = dt$ ⁠.

Why?

$\sin x$ appears in the denominator, and the numerator $\cos x$ is its derivative. So $t = \sin x$ leaves a rational function of $t$ only.

When $x = 0$ ⁠, $t = 0$ ⁠; when $x = \frac{\pi}{2}$ ⁠, $t = 1$ ⁠.

I = \int_0^1 \frac{dt}{(1 + t)(4 + t)}

Why?

The limits change with the variable: $\sin 0 = 0$ and $\sin\frac{\pi}{2} = 1$ ⁠.

Watch out: with the new limits in $t$ ⁠, there is no need to go back to $x$ at the end.

Let $\frac{1}{(1 + t)(4 + t)} = \frac{A}{1 + t} + \frac{B}{4 + t}$ ⁠.

\Rightarrow 1 = A(4 + t) + B(1 + t)

Putting $t = -1$ ⁠: $A = \frac13$ ⁠. Putting $t = -4$ ⁠: $B = -\frac13$ ⁠.

Why?

The denominator has two different linear factors, so it splits into two simple fractions. Multiplying by $(1 + t)(4 + t)$ gives the identity above.

Choose values of $t$ that make one bracket zero. At $t = -1$ ⁠: $1 = 3A$ ⁠. At $t = -4$ ⁠: $1 = -3B$ ⁠.

\begin{aligned} t = -1&:\ 1 = A(4 - 1) + B(1 - 1) \\ &\quad 1 = 3A + 0 \\ &\quad A = \frac13 \\[4pt] t = -4&:\ 1 = A(4 - 4) + B(1 - 4) \\ &\quad 1 = 0 - 3B \\ &\quad B = -\frac13 \end{aligned}

\begin{aligned} I &= \frac13\left[\int_0^1 \frac{dt}{1 + t} - \int_0^1 \frac{dt}{4 + t}\right] \end{aligned}

Why?

Put $A = \frac13$ and $B = -\frac13$ into the split:

\begin{aligned} &\frac{1}{(1 + t)(4 + t)} \\ &= \frac{1/3}{1 + t} + \frac{-1/3}{4 + t} \\ &= \frac13\left[\frac{1}{1 + t} - \frac{1}{4 + t}\right] \end{aligned}

The common factor $\frac13$ comes outside the integral.

\begin{aligned} I &= \frac13\Big[\log(1 + t) - \log(4 + t)\Big]_0^1 \\ &= \frac13[(\log 2 - \log 5) \\ &\qquad - (\log 1 - \log 4)] \\ &= \frac13[\log 2 - \log 5 + \log 4] \end{aligned}

Why?

Each piece is a standard log integral:

\int \frac{dt}{a + t} = \log|a + t| + C

Both $1 + t$ and $4 + t$ are positive for $0 \le t \le 1$ ⁠, so no modulus is needed.

Upper limit minus lower limit, with $\log 1 = 0$ ⁠. Then combine the logs:

\begin{aligned} t = 1&:\ \log(1 + 1) - \log(4 + 1) \\ &\quad = \log 2 - \log 5 \\ t = 0&:\ \log(1 + 0) - \log(4 + 0) \\ &\quad = \log 1 - \log 4 = -\log 4 \end{aligned}

\begin{aligned} &(\log 2 - \log 5) - (-\log 4) \\ &= \log 2 - \log 5 + \log 4 \end{aligned}

\begin{aligned} \log 2 + \log 4 &= \log(2 \times 4) = \log 8 \\ \log 8 - \log 5 &= \log\frac{8}{5} \end{aligned}

$\therefore\ I = \dfrac13\log\dfrac85$

Where this comes from

Chapter 7, Integrals: Integration by substitution; partial fractions.

Question type: (a) Substitution; (b) Substitution + Partial fractions (combined)

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: Two roots of $x$ sit at the bottom, $\sqrt x$ and $\sqrt[3]x$ ⁠. Putting $x = t^6$ turns them into $t^3$ and $t^2$ ⁠, and $dx$ into $6t^5\,dt$ ⁠.

(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: First, the $\cos x$ on top is the derivative of the $\sin x$ at the bottom: put $\sin x = t$ ⁠, and the limits $0$ and $\frac{\pi}{2}$ become $0$ and $1$ ⁠.

(b) Type 7. Partial fractions: break a fraction into simple pieces

Spot it: a fraction of two polynomials whose bottom splits into factors.

Here: The fraction left in $t$ has the bottom $(1 + t)(4 + t)$ ⁠: split it into $\frac{A}{1 + t} + \frac{B}{4 + t}$ ⁠, with $A = \frac13$ and $B = -\frac13$ ⁠.

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Q9: $\int \frac{dx}{\sqrt x + \sqrt[3]x}$ ⁠, or $\int_0^{\pi/2} \frac{\cos x}{(1 + \sin x)(4 + \sin x)}\,dx$