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CBSE Board Examination 2022 (Term 2) · Section A · 2 marks

Q3: F(x)F(x) from F′(x)=sec⁡4xcosec⁡4xF'(x) = \frac{\sec^4x}{\operatorname{cosec}^4x}⁠, or ∫log⁡x−3(log⁡x)4 dx\int \frac{\log x - 3}{(\log x)^4}\,dx

Board 2022 T2Ch 7 · Integrals
The question

(a) If ddx[F(x)]=sec⁡4xcosec⁡4x\dfrac{d}{dx}[F(x)] = \dfrac{\sec^4x}{\operatorname{cosec}^4x} and F(π4)=π4F\left(\dfrac{\pi}{4}\right) = \dfrac{\pi}{4}⁠, then find F(x)F(x)⁠.

OR

(b) Find : ∫log⁡x−3(log⁡x)4 dx\displaystyle\int \frac{\log x - 3}{(\log x)^4}\,dx⁠.

Exam answer

(a)

sec⁡4xcosec⁡4x=sin⁡4xcos⁡4x=tan⁡4x\frac{\sec^4x}{\operatorname{cosec}^4x} = \frac{\sin^4x}{\cos^4x} = \tan^4x
Why?

We know F′(x)F'(x)⁠, so F(x)F(x) is its integral. First make the integrand simpler. Since sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x} and cosec⁡x=1sin⁡x\operatorname{cosec} x = \frac{1}{\sin x}⁠,

sec⁡4xcosec⁡4x=1cos⁡4x×sin⁡4x\frac{\sec^4x}{\operatorname{cosec}^4x} = \frac{1}{\cos^4x} \times \sin^4x

and sin⁡xcos⁡x=tan⁡x\frac{\sin x}{\cos x} = \tan x⁠.

1cos⁡4x×sin⁡4x=sin⁡4xcos⁡4x=(sin⁡xcos⁡x)4=tan⁡4x\begin{aligned} \frac{1}{\cos^4x} \times \sin^4x &= \frac{\sin^4x}{\cos^4x} \\ &= \left(\frac{\sin x}{\cos x}\right)^4 \\ &= \tan^4x \end{aligned}
F(x)=∫tan⁡4x dx=∫tan⁡2x (sec⁡2x−1) dx\begin{aligned} F(x) &= \int \tan^4x\,dx \\ &= \int \tan^2x\,(\sec^2x - 1)\,dx \end{aligned}
Why?

There is no standard integral for tan⁡4x\tan^4x⁠. Write it as tan⁡2x⋅tan⁡2x\tan^2x \cdot \tan^2x and change only one factor, using

1+tan⁡2x=sec⁡2x1 + \tan^2x = \sec^2x

so tan⁡2x=sec⁡2x−1\tan^2x = \sec^2x - 1⁠:

tan⁡4x=tan⁡2x⋅tan⁡2x=tan⁡2x (sec⁡2x−1)\begin{aligned} \tan^4x &= \tan^2x \cdot \tan^2x \\ &= \tan^2x\,(\sec^2x - 1) \end{aligned}

The piece tan⁡2xsec⁡2x\tan^2x\sec^2x is then a power of tan⁡x\tan x times its derivative. The leftover −tan⁡2x-\tan^2x is replaced in the same way, giving −sec⁡2x+1-\sec^2x + 1⁠.

=∫(tan⁡2xsec⁡2x−tan⁡2x) dx=∫(tan⁡2xsec⁡2x−sec⁡2x+1) dx\begin{aligned} &= \int (\tan^2x\sec^2x - \tan^2x)\,dx \\ &= \int (\tan^2x\sec^2x \\ &\qquad - \sec^2x + 1)\,dx \end{aligned}
Why?

Multiply out the bracket:

tan⁡2x (sec⁡2x−1)=tan⁡2xsec⁡2x−tan⁡2x\begin{aligned} &\tan^2x\,(\sec^2x - 1) \\ &= \tan^2x\sec^2x - \tan^2x \end{aligned}

Then change the last tan⁡2x\tan^2x to sec⁡2x−1\sec^2x - 1⁠:

−tan⁡2x=−(sec⁡2x−1)=−sec⁡2x+1\begin{aligned} -\tan^2x &= -(\sec^2x - 1) \\ &= -\sec^2x + 1 \end{aligned}

For the first term put tan⁡x=t\tan x = t⁠, sec⁡2x dx=dt\sec^2x\,dx = dt⁠.

∫tan⁡2xsec⁡2x dx=∫t2 dt=t33=tan⁡3x3\begin{aligned} \int \tan^2x\sec^2x\,dx &= \int t^2\,dt \\ &= \frac{t^3}{3} = \frac{\tan^3x}{3} \end{aligned} F(x)=tan⁡3x3−tan⁡x+x+CF(x) = \frac{\tan^3x}{3} - \tan x + x + C
Why?

In the first term, ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2x⁠, so with t=tan⁡xt = \tan x the factor sec⁡2x dx\sec^2x\,dx is exactly dtdt⁠. Then the power rule:

∫t2 dt=t2+12+1=t33\int t^2\,dt = \frac{t^{2 + 1}}{2 + 1} = \frac{t^3}{3}

The other two terms are standard integrals:

∫sec⁡2x dx=tan⁡x∫1 dx=x\begin{aligned} \int \sec^2x\,dx &= \tan x \\ \int 1\,dx &= x \end{aligned}

Put the three pieces together, keeping the signs:

∫tan⁡2xsec⁡2x dx=tan⁡3x3−∫sec⁡2x dx=−tan⁡x+∫1 dx=+x\begin{aligned} &\int \tan^2x\sec^2x\,dx = \frac{\tan^3x}{3} \\ &-\int \sec^2x\,dx = -\tan x \\ &+\int 1\,dx = +x \end{aligned}

One constant CC covers all three terms.

Given F(π4)=π4F\left(\frac{\pi}{4}\right) = \frac{\pi}{4}⁠:

13−1+π4+C=π4⇒C=23\frac13 - 1 + \frac{\pi}{4} + C = \frac{\pi}{4} \Rightarrow C = \frac23
Why?

Put x=π4x = \frac{\pi}{4} in F(x)F(x)⁠. Since tan⁡π4=1\tan\frac{\pi}{4} = 1⁠, the first term is 133=13\frac{1^3}{3} = \frac13 and the second is −1-1⁠.

F(π4)=(1)33−1+π4+C=13−1+π4+C\begin{aligned} F\left(\frac{\pi}{4}\right) &= \frac{(1)^3}{3} - 1 + \frac{\pi}{4} + C \\ &= \frac13 - 1 + \frac{\pi}{4} + C \end{aligned}

The π4\frac{\pi}{4} cancels from both sides, leaving C=1−13=23C = 1 - \frac13 = \frac23⁠.

13−1+π4+C=π413−1+C=0C=1−13=33−13=23\begin{aligned} \frac13 - 1 + \frac{\pi}{4} + C &= \frac{\pi}{4} \\ \frac13 - 1 + C &= 0 \\ C &= 1 - \frac13 \\ &= \frac33 - \frac13 = \frac23 \end{aligned}

∴ F(x)=tan⁡3x3−tan⁡x+x+23\therefore\ F(x) = \frac{\tan^3x}{3} - \tan x + x + \frac23

(b)

Let I=∫log⁡x−3(log⁡x)4 dxI = \displaystyle\int \frac{\log x - 3}{(\log x)^4}\,dx⁠.

Put log⁡x=t\log x = t⁠, i.e. x=etx = e^t⁠. Then dx=et dtdx = e^t\,dt⁠.

Why?

Everything in the integrand depends on log⁡x\log x⁠, so t=log⁡xt = \log x turns it into a simple fraction in tt⁠.

From x=etx = e^t⁠, dxdt=et\frac{dx}{dt} = e^t⁠. This is where the factor ete^t in the next line comes from.

log⁡x−3(log⁡x)4 dx=t−3t4 et dt\begin{aligned} &\frac{\log x - 3}{(\log x)^4}\,dx \\ &= \frac{t - 3}{t^4}\,e^t\,dt \end{aligned}
I=∫t−3t4 et dt=∫(1t3−3t4)et dt\begin{aligned} I &= \int \frac{t - 3}{t^4}\,e^t\,dt \\ &= \int \left(\frac{1}{t^3} - \frac{3}{t^4}\right)e^t\,dt \end{aligned}
Why?

Split the fraction term by term: tt4=1t3\frac{t}{t^4} = \frac{1}{t^3} and the other term is −3t4-\frac{3}{t^4}⁠.

t−3t4=tt4−3t4=1t3−3t4\begin{aligned} \frac{t - 3}{t^4} &= \frac{t}{t^4} - \frac{3}{t^4} \\ &= \frac{1}{t^3} - \frac{3}{t^4} \end{aligned}

Now notice that the second term is the derivative of the first:

ddt(t−3)=−3t−3−1=−3t−4=−3t4\begin{aligned} \frac{d}{dt}\left(t^{-3}\right) &= -3t^{-3 - 1} \\ &= -3t^{-4} = -\frac{3}{t^4} \end{aligned}

This is ∫et[f(t)+f′(t)] dt\int e^t[f(t) + f'(t)]\,dt with f(t)=1t3f(t) = \frac{1}{t^3}⁠, f′(t)=−3t4f'(t) = -\frac{3}{t^4}⁠.

Using ∫et[f(t)+f′(t)] dt\int e^t[f(t) + f'(t)]\,dt =etf(t)+C= e^tf(t) + C⁠,

I=ett3+CI = \frac{e^t}{t^3} + C
Why?

The standard result comes from the product rule:

ddt[etf(t)]=etf(t)+etf′(t)\begin{aligned} &\frac{d}{dt}\left[e^tf(t)\right] \\ &= e^tf(t) + e^tf'(t) \end{aligned}

Here etf(t)=et×1t3=ett3e^tf(t) = e^t \times \frac{1}{t^3} = \frac{e^t}{t^3}⁠.

To return to xx⁠, use et=xe^t = x and t=log⁡xt = \log x⁠.

ett3=x(log⁡x)3\frac{e^t}{t^3} = \frac{x}{(\log x)^3}

∴ I=x(log⁡x)3+C\therefore\ I = \dfrac{x}{(\log x)^3} + C

Where this comes from

Chapter 7, Integrals: Integrals: tan⁴x; the eᵗ[f(t) + f′(t)] form.

Question type: (a) Substitution; (b) Substitution + The exe^x shortcut (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\log x and 1x\frac{1}{x}⁠, or x2+1x^2 + 1 and xx⁠.

Here: Writing tan⁡4x=tan⁡2x(sec⁡2x−1)\tan^4x = \tan^2x(\sec^2x - 1) makes tan⁡2xsec⁡2x\tan^2x\sec^2x appear: tan⁡x\tan x with its derivative sec⁡2x\sec^2x beside it. So put tan⁡x=t\tan x = t⁠; the rest is ∫(1−sec⁡2x) dx\int(1 - \sec^2x)\,dx⁠.

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

Spot it: a function and its derivative both appear, like log⁡x\log x and 1x\frac{1}{x}⁠, or x2+1x^2 + 1 and xx⁠.

Here: Everything here is a function of log⁡x\log x⁠, so first put log⁡x=t\log x = t⁠. Then x=etx = e^t and dx=et dtdx = e^t\,dt⁠, which brings in an ete^t⁠.

(b) Type 9. The exe^x shortcut: a function and its derivative in a bracket

Spot it: exe^x times a bracket of two parts, where one part is the derivative of the other.

Here: The fraction left in tt splits into 1t3−3t4\frac{1}{t^3} - \frac{3}{t^4}⁠, and −3t4-\frac{3}{t^4} is the derivative of 1t3\frac{1}{t^3}⁠. So f(t)=1t3f(t) = \frac{1}{t^3} and the answer is ett3+C\frac{e^t}{t^3} + C⁠.

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