NCERT Exemplar Solutions Class 12 Maths Integrals — Short Answer Q18–34

NCERT Exemplar Problems, Class 12 Mathematics — Integrals, Short Answer questions 18 to 34.

The second half of the Short Answer band changes character twice. Questions 18 to 26 are still indefinite integrals, but the algebra is heavier — partial fractions on quartics, and two substitutions the book has to hint at. Questions 27 and 28 then ask for a limit of sums, which is a different skill from everything around it. From 29 onwards every question is a definite integral, and several are solved faster by a property than by finding an antiderivative at all.

Key insight. From Question 29 on, ask what the limits can do for you before you start integrating. Question 33 falls out of $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ in three lines and is a long, unpleasant integration by parts if you do not spot it. Question 31’s answer is $\pi$ with no antiderivative evaluated at all once the square is completed. The exemplar puts these together precisely to train that reflex.

Question 18

$$\int\frac{x}{x^4-1}\,dx$$

Solution. The numerator is $x$ and the denominator is a function of $x^2$, so the substitution writes itself: $t = x^2$, $dt = 2x\,dx$.

$$\frac{1}{2}\int\frac{dt}{t^2-1} = \frac{1}{2}\cdot\frac{1}{2}\log\left|\frac{t-1}{t+1}\right| + C$$

Why not partial fractions in $x$. Factorising $x^4-1$ into $(x-1)(x+1)(x^2+1)$ gives three fractions and a $\tan^{-1}$ term that must ultimately cancel. Substituting first keeps it to one standard form.

Question 19

$$\int\frac{x^2}{1-x^4}\,dx \qquad \textbf{(put } x^2 = t\textbf{)}$$

Solution. Here the numerator is $x^2$, not $x$, so the hinted substitution does not clear the $dx$ — and that is the point of the question. Split into partial fractions in $x^2$ instead. Noting that

$$\frac{1}{1-x^2} – \frac{1}{1+x^2} = \frac{(1+x^2)-(1-x^2)}{1-x^4} = \frac{2x^2}{1-x^4}$$

gives $\dfrac{x^2}{1-x^4} = \dfrac12\left[\dfrac{1}{1-x^2} – \dfrac{1}{1+x^2}\right]$, and both pieces are standard:

$$\frac{1}{2}\left[\frac{1}{2}\log\left|\frac{1+x}{1-x}\right| – \tan^{-1}x\right] + C$$

Compare with Question 18. Same denominator shape, one more power of $x$ on top, and a completely different method. The test is whether $x\,dx$ is available to become $dt$ — in 18 it is, in 19 it is not.

Question 20

$$\int\sqrt{2ax-x^2}\,dx$$

Solution. Complete the square, remembering that the $x^2$ carries a minus sign:

$$2ax – x^2 = -\left(x^2-2ax\right) = a^2 – (x-a)^2$$

With $u = x-a$ this is $\int\sqrt{a^2-u^2}\,du$, the standard form

$$\int\sqrt{a^2-u^2}\,du = \frac{u}{2}\sqrt{a^2-u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} + C$$

Question 21

$$\int\frac{\sin^{-1}x}{\left(1-x^2\right)^{\frac{3}{2}}}\,dx$$

Solution. Substitute $t = \sin^{-1}x$, so $x = \sin t$, $dx = \cos t\,dt$ and $\left(1-x^2\right)^{3/2} = \cos^3 t$:

$$\int\frac{t\cos t}{\cos^3 t}\,dt = \int t\sec^2 t\,dt$$

which is by parts with $t$ as the first function:

$$t\tan t – \int\tan t\,dt = t\tan t + \log|\cos t| + C$$

Returning to $x$, where $\tan t = \dfrac{x}{\sqrt{1-x^2}}$ and $\cos t = \sqrt{1-x^2}$:

Why the substitution comes before the parts. Attempting parts directly on $\sin^{-1}x \cdot (1-x^2)^{-3/2}$ needs the antiderivative of $(1-x^2)^{-3/2}$ first, which is itself $\frac{x}{\sqrt{1-x^2}}$ — findable, but the substitution turns the whole thing into the familiar $\int t\sec^2 t\,dt$ in one step.

Question 22

$$\int\frac{\left(\cos 5x + \cos 4x\right)}{1-2\cos 3x}\,dx$$

Solution. This looks impossible and is a factorisation. The claim is that

$$\cos 5x + \cos 4x = -\left(\cos x + \cos 2x\right)\left(1-2\cos 3x\right)$$

and it is worth expanding the right side to see why, rather than taking it on trust. Multiply out and use $2\cos A\cos B = \cos(A+B) + \cos(A-B)$:

$$2\cos 3x\cos x = \cos 4x + \cos 2x, \qquad 2\cos 3x\cos 2x = \cos 5x + \cos x$$

so the right side is $-\cos x – \cos 2x + \cos 4x + \cos 2x + \cos 5x + \cos x = \cos 4x + \cos 5x$, as claimed. The denominator therefore cancels entirely:

$$\int -\left(\cos x + \cos 2x\right)dx = -\sin x – \frac{\sin 2x}{2} + C$$

How you would find this without being told. Convert the numerator with the sum-to-product formula: $\cos 5x + \cos 4x = 2\cos\frac{9x}{2}\cos\frac{x}{2}$. Writing $1-2\cos 3x$ in half-angle form and cancelling leads to the same place. But the fastest route in an examination is to suspect that the denominator divides the numerator exactly, and check.

Question 23

$$\int\frac{\sin^6 x + \cos^6 x}{\sin^2 x\,\cos^2 x}\,dx$$

Solution. Use the identity for a sum of cubes on $\left(\sin^2x\right)^3 + \left(\cos^2x\right)^3$. Since $\sin^2x + \cos^2x = 1$,

$$\sin^6 x + \cos^6 x = 1 – 3\sin^2 x\cos^2 x$$

$$\frac{1-3\sin^2x\cos^2x}{\sin^2x\cos^2x} = \frac{1}{\sin^2x\cos^2x} – 3$$

and the first term splits neatly, because $1 = \sin^2x + \cos^2x$:

$$\frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x} = \sec^2x + \operatorname{cosec}^2x$$

$$\int\left(\sec^2x + \operatorname{cosec}^2x – 3\right)dx = \tan x – \cot x – 3x + C$$

The move worth keeping. Writing $1$ as $\sin^2x+\cos^2x$ to split a fraction is one of the most useful tricks in trigonometric integration, and it appears again in the Long Answer band.

Question 24

$$\int\frac{\sqrt{x}}{\sqrt{a^3-x^3}}\,dx$$

Solution. The root contains $x^3$ and the numerator carries $x^{1/2}$, and $\frac{d}{dx}\left(x^{3/2}\right) = \frac32 x^{1/2}$ — so $t = x^{3/2}$ makes the numerator into $dt$ exactly.

With $dt = \frac32\sqrt{x}\,dx$ and $x^3 = t^2$:

$$\frac{2}{3}\int\frac{dt}{\sqrt{\left(a^{3/2}\right)^2-t^2}} = \frac{2}{3}\sin^{-1}\frac{t}{a^{3/2}} + C$$

How to see the substitution. Whenever a root holds $a^n – x^n$ and the numerator is $x^{n/2-1}$, the substitution $t = x^{n/2}$ turns it into $\int\frac{dt}{\sqrt{A^2-t^2}}$. Here $n=3$, so $x^{1/2}$ on top is exactly what is needed.

Question 25

$$\int\frac{\cos x – \cos 2x}{1-\cos x}\,dx$$

Solution. Another cancelling factorisation. Put $\cos 2x = 2\cos^2x-1$ in the numerator:

$$\cos x – \left(2\cos^2 x – 1\right) = -2\cos^2x + \cos x + 1 = \left(2\cos x + 1\right)\left(1-\cos x\right)$$

(check by expanding: $2\cos x – 2\cos^2x + 1 – \cos x = -2\cos^2x + \cos x + 1$). The $\left(1-\cos x\right)$ cancels with the denominator:

$$\int\left(2\cos x + 1\right)dx = 2\sin x + x + C$$

Why $\cos 2x = 2\cos^2x – 1$ and not one of the other two forms. The denominator is in $\cos x$, so the numerator must be too. Choosing the $1-2\sin^2x$ form leaves a mix that will not factor against $1-\cos x$.

Question 26

$$\int\frac{dx}{x\sqrt{x^4-1}} \qquad \textbf{(Hint: put } x^2 = \sec\theta\textbf{)}$$

Solution. With $x^2 = \sec\theta$ we get $2x\,dx = \sec\theta\tan\theta\,d\theta$, so

$$\frac{dx}{x} = \frac{2x\,dx}{2x^2} = \frac{\sec\theta\tan\theta\,d\theta}{2\sec\theta} = \frac{\tan\theta}{2}\,d\theta$$

and the root becomes $\sqrt{\sec^2\theta-1} = \tan\theta$. Everything cancels:

$$\int\frac{\tan\theta\,d\theta}{2\tan\theta} = \frac{\theta}{2} + C$$

Why the hint is needed. $\sqrt{x^4-1}$ is the $\sqrt{u^2-a^2}$ shape, and that shape always wants a secant substitution. The hint tells you which power of $x$ to set equal to $\sec\theta$, which is the only real decision.

Question 27

Evaluate as a limit of sums: $\displaystyle\int_0^2\left(x^2+3\right)dx$

Solution. The definition, with $a=0$, $b=2$, $h = \dfrac{b-a}{n} = \dfrac{2}{n}$:

$$\int_a^b f(x)\,dx = \lim_{n\to\infty} h\left[f(a) + f(a+h) + \cdots + f\left(a+(n-1)h\right)\right]$$

Here $f(a+rh) = \left(\frac{2r}{n}\right)^2 + 3 = \frac{4r^2}{n^2} + 3$, so the sum is

$$\frac{2}{n}\sum_{r=0}^{n-1}\left(\frac{4r^2}{n^2}+3\right) = \frac{8}{n^3}\sum_{r=0}^{n-1}r^2 + \frac{2}{n}\cdot 3n = \frac{8}{n^3}\cdot\frac{(n-1)n(2n-1)}{6} + 6$$

As $n\to\infty$ the first term tends to $\dfrac{8}{6}\cdot 2 = \dfrac{8}{3}$.

Why you may not simply use the antiderivative. The question names the method. Writing $\left[\frac{x^3}{3}+3x\right]_0^2$ gets the same number and answers a different question — the point is to show the definite integral is that limit, which is what the Fundamental Theorem then lets you shortcut.

A note on the current syllabus. Questions 27 and 28 ask for a technique the present NCERT textbook no longer drills. The rationalised book still defines the definite integral as the limit of a sum, and says in as many words that the Fundamental Theorem “gives us a method of calculating the definite integral more easily, without calculating the limit of a sum” — but it sets no exercise on it. The exemplar dates from before that change. It is still worth doing: the method is on the JEE syllabus, and it is the only place the definition is ever used for anything.

Question 28

Evaluate as a limit of sums: $\displaystyle\int_0^2 e^x\,dx$

Solution. Same definition, $h = \frac{2}{n}$, and the sum is now geometric:

$$h\sum_{r=0}^{n-1}e^{rh} = h\cdot\frac{e^{nh}-1}{e^h-1} = \frac{2}{n}\cdot\frac{e^2-1}{e^{2/n}-1}$$

using $nh = 2$. As $n\to\infty$, write $h=2/n\to 0$ and use $\lim_{h\to 0}\dfrac{h}{e^{h}-1} = 1$:

$$\lim_{n\to\infty}\frac{2/n}{e^{2/n}-1}\left(e^2-1\right) = 1\cdot\left(e^2-1\right)$$

The one limit to know. $\dfrac{e^h-1}{h}\to 1$ as $h\to 0$ is what makes every exponential limit-of-sums question finish. It is the same limit that defines the derivative of $e^x$ at $0$.

Question 29

$$\int_0^1\frac{dx}{e^x+e^{-x}}$$

Solution. Multiply above and below by $e^x$ to clear the negative exponent:

$$\int_0^1\frac{e^x\,dx}{e^{2x}+1}$$

Now $u = e^x$, $du = e^x dx$, and the limits become $u: 1 \to e$:

$$\int_1^{e}\frac{du}{u^2+1} = \left[\tan^{-1}u\right]_1^{e} = \tan^{-1}e – \tan^{-1}1$$

Changing the limits with the variable. Because $u = e^x$, the limits $0$ and $1$ become $1$ and $e$. Substituting back to $x$ at the end and then applying the original limits is equally valid and one step longer.

Question 30

$$\int_0^{\frac{\pi}{2}}\frac{\tan x\,dx}{1+m^2\tan^2 x}$$

Solution. Write everything over $\cos^2 x$ and substitute. With $\tan x = t$, $\sec^2x\,dx = dt$, the integrand becomes

$$\frac{\tan x}{1+m^2\tan^2x}\,dx = \frac{t}{\left(1+m^2t^2\right)\left(1+t^2\right)}\,dt$$

and $x: 0\to\frac{\pi}{2}$ gives $t: 0\to\infty$. Now put $t^2 = z$, $2t\,dt = dz$:

$$\frac{1}{2}\int_0^{\infty}\frac{dz}{\left(1+m^2z\right)\left(1+z\right)}$$

Partial fractions give $\dfrac{1}{(1+m^2z)(1+z)} = \dfrac{1}{1-m^2}\left[\dfrac{1}{1+z} – \dfrac{m^2}{1+m^2z}\right]$, and the two logarithms combine before the limit is taken:

$$\frac{1}{2\left(1-m^2\right)}\left[\log\frac{1+z}{1+m^2z}\right]_0^{\infty} = \frac{1}{2\left(1-m^2\right)}\log\frac{1}{m^2} = \frac{\log m}{m^2-1}$$

Why the two logs must be combined before letting $z\to\infty$. Separately each diverges; together the ratio tends to $\frac{1}{m^2}$, which is finite. Splitting them and taking limits one at a time gives $\infty-\infty$.

Question 31

$$\int_1^2\frac{dx}{\sqrt{(x-1)(2-x)}}$$

Solution. Expand and complete the square:

$$(x-1)(2-x) = -x^2+3x-2 = \frac{1}{4} – \left(x-\frac{3}{2}\right)^2$$

$$\int_1^2\frac{dx}{\sqrt{\left(\frac12\right)^2-\left(x-\frac32\right)^2}} = \left[\sin^{-1}\!\left(\frac{x-\frac32}{\frac12}\right)\right]_1^2 = \sin^{-1}(1) – \sin^{-1}(-1)$$

Why the answer is $\pi$ and not something messier. The limits are exactly the roots of the quadratic, so the substituted variable runs over the whole of $[-1,1]$ — and $\sin^{-1}$ spans its full range. Any integral of this form between the two roots gives $\pi$, whatever the roots are.

Question 32

$$\int_0^1\frac{x\,dx}{\sqrt{1+x^2}}$$

Solution. The numerator is half the derivative of $1+x^2$, so no substitution is really needed:

$$\int_0^1\frac{x\,dx}{\sqrt{1+x^2}} = \left[\sqrt{1+x^2}\right]_0^1$$

Question 33

$$\int_0^{\pi}x\sin x\cos^2 x\,dx$$

Solution. Integration by parts works and is long. Use the property $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ instead, with $a=\pi$:

$$I = \int_0^{\pi}(\pi-x)\sin(\pi-x)\cos^2(\pi-x)\,dx = \int_0^{\pi}(\pi-x)\sin x\cos^2x\,dx$$

since $\sin(\pi-x) = \sin x$ and $\cos^2(\pi-x) = \cos^2x$. Adding this to the original:

$$2I = \int_0^{\pi}\pi\sin x\cos^2x\,dx = \pi\left[-\frac{\cos^3x}{3}\right]_0^{\pi} = \pi\left(\frac13+\frac13\right) = \frac{2\pi}{3}$$

This is the question the band is built around. The $x$ that makes parts necessary is exactly what the property removes. Whenever you see a lone $x$ multiplying a function that is symmetric about the midpoint of the interval, try the property first.

Question 34

$$\int_0^{\frac{1}{2}}\frac{dx}{\left(1+x^2\right)\sqrt{1-x^2}} \qquad \textbf{(Hint: let } x=\sin\theta\textbf{)}$$

Solution. With $x=\sin\theta$, $dx=\cos\theta\,d\theta$ and $\sqrt{1-x^2}=\cos\theta$, the root cancels:

$$\int_0^{\pi/6}\frac{d\theta}{1+\sin^2\theta}$$

Divide above and below by $\cos^2\theta$ to reach a $\tan$ substitution:

$$\int_0^{\pi/6}\frac{\sec^2\theta\,d\theta}{\sec^2\theta+\tan^2\theta} = \int_0^{\pi/6}\frac{\sec^2\theta\,d\theta}{1+2\tan^2\theta}$$

Now $u=\tan\theta$, $du=\sec^2\theta\,d\theta$, with $u: 0\to\frac{1}{\sqrt3}$:

$$\int_0^{1/\sqrt3}\frac{du}{1+2u^2} = \frac{1}{\sqrt2}\left[\tan^{-1}\left(\sqrt2\,u\right)\right]_0^{1/\sqrt3}$$

Two substitutions, and both are forced. The first clears the root; the second turns $1+\sin^2\theta$ into a quadratic in $\tan\theta$. Neither is optional, and recognising that $\sec^2 = 1+\tan^2$ lets the denominator collapse is the step that makes the second one work.

Common mistakes

• Using the hinted substitution in Question 19 because it is printed. $x^2=t$ clears the denominator but leaves $x\,dx$ unavailable, so it cannot finish. Contrast Question 18, where the same substitution is exactly right — the difference is one power of $x$ in the numerator.
• Evaluating Questions 27 and 28 with the Fundamental Theorem. The question says as a limit of sums. The antiderivative gives the right number and no marks — and since the current textbook sets no exercise on the method, this is the pair most likely to be met for the first time here.
• Splitting the two logarithms in Question 30 before taking the limit. Each diverges alone; only their ratio is finite.
• Reaching for parts in Question 33. It works and takes five times as long. The property $\int_0^a f(x) = \int_0^a f(a-x)$ removes the $x$ entirely.
• Choosing the wrong form of $\cos 2x$ in Question 25. The denominator is in $\cos x$, so the numerator must become a polynomial in $\cos x$ — which only $2\cos^2x-1$ achieves.
• Forgetting to move the limits in Questions 29, 30 and 34. Every one of them substitutes, and each substitution changes the limits. Carrying the old ones through is the single most common slip in the definite half of this band.
• Abandoning Question 22 as impossible. The denominator divides the numerator exactly. A fraction with a trigonometric polynomial above and below is always worth testing for a cancelling factor before anything else.

Practise next

• Exemplar Long Answer Q35–47, where the partial fractions of Questions 18 and 19 return on quartics and the definite-integral properties used here are pushed much harder.
• Exemplar Short Answer Q1–17, the first half of this band, if the substitutions and standard forms above felt unfamiliar.
• NCERT textbook Exercise 7.9 for definite integrals by substitution, which Questions 29, 30 and 34 all depend on.
• NCERT textbook Exercise 7.10 for the properties of definite integrals that Question 33 turns on — it is the textbook exercise closest to the back half of this page.