NCERT Exemplar Solutions Class 12 Maths Integrals — Short Answer Q1–17

NCERT Exemplar Problems, Class 12 Mathematics — Integrals, Short Answer questions 1 to 17.

The exemplar’s Short Answer band for Integrals is 34 questions long, and this page solves the first seventeen. They are not arranged by difficulty but by technique, and knowing which technique a question is testing is most of the work:

• Questions 1 and 2 are verifications, not evaluations — the answer is printed in the question and you have to show it is right.
• Questions 3, 4 and 10 reduce to a polynomial division before anything else happens.
• Questions 5, 7, 8, 9 and 12 are substitutions, three of them hidden inside a trigonometric identity.
• Questions 13 to 17 are the standard forms — $\dfrac{1}{\sqrt{a^2-x^2}}$, $\dfrac{1}{\sqrt{x^2+a^2}}$ and $\sqrt{x^2+a^2}$ — reached by completing the square.

Key insight. Before integrating anything, ask whether the numerator is the derivative of the denominator, or can be made so. Questions 2, 5 and 16 all collapse the moment you notice it, and Question 16 is the one worth studying: $3x-1$ is not the derivative of $x^2+9$, but $3x$ is a multiple of it, so splitting the fraction into a part that is and a part that is not turns one hard integral into two standard ones. That split is the single most reusable move in this band.

Question 1

Verify that $\displaystyle\int\frac{2x-1}{2x+3}\,dx = x – \log\left|(2x+3)^2\right| + C$.

Solution. A verification can be done two ways: differentiate the right-hand side, or integrate the left. Integrating is more instructive here, because it shows where the squared bracket comes from.

The numerator is one degree in $x$ and so is the denominator, so force the denominator into the numerator:

$$2x-1 = (2x+3) – 4 \qquad\Longrightarrow\qquad \frac{2x-1}{2x+3} = 1 – \frac{4}{2x+3}$$

$$\int\frac{2x-1}{2x+3}\,dx = \int 1\,dx – 4\int\frac{dx}{2x+3} = x – 4\cdot\frac{1}{2}\log|2x+3| + C = x – 2\log|2x+3| + C$$

Now $2\log|2x+3| = \log|2x+3|^2 = \log\left|(2x+3)^2\right|$, which is exactly what the question claims.

Why the book writes it that way. The squared form looks stranger than $-2\log|2x+3|$, and it is there to test whether you recognise $n\log a = \log a^n$ under the modulus. A student who differentiates instead gets $1 – \frac{4}{2x+3}$ and has to work backwards to see the agreement.

Question 2

Verify that $\displaystyle\int\frac{2x+3}{x^2+3x}\,dx = \log\left|x^2+3x\right| + C$.

Solution. Look at the denominator and differentiate it:

$$\frac{d}{dx}\left(x^2+3x\right) = 2x+3$$

which is the numerator exactly. So the integral is of the form $\int \dfrac{f'(x)}{f(x)}\,dx$, whose value is $\log|f(x)| + C$ with no substitution needed.

Why questions 1 and 2 have no answer in the key. They are the only two questions in the whole unit that NCERT’s answer key skips, and it is not an omission — “verify the following” states the answer in the question, exactly as a “prove that” question does in the textbook. There is nothing left for a key to print.

Question 3

$$\int\frac{\left(x^2+2\right)dx}{x+1}$$

Solution. The numerator’s degree is higher than the denominator’s, so divide first. Nothing else can usefully happen until it is a polynomial plus a proper fraction.

$$x^2+2 = (x+1)(x-1) + 3 \qquad\Longrightarrow\qquad \frac{x^2+2}{x+1} = x – 1 + \frac{3}{x+1}$$

$$\int\left(x – 1 + \frac{3}{x+1}\right)dx = \frac{x^2}{2} – x + 3\log|x+1| + C$$

Where the 3 comes from. Substituting $x = -1$ into $x^2 + 2$ gives $3$, which is the remainder. That is quicker than long division and worth doing in your head.

Question 4

$$\int\frac{e^{6\log x} – e^{5\log x}}{e^{4\log x} – e^{3\log x}}\,dx$$

Solution. Every term is of the form $e^{n\log x}$, and $e^{n\log x} = \left(e^{\log x}\right)^n = x^n$. Rewrite the whole thing as powers before integrating:

$$\frac{x^6 – x^5}{x^4 – x^3} = \frac{x^5(x-1)}{x^3(x-1)} = x^2 \qquad (x \neq 1)$$

$$\int x^2\,dx = \frac{x^3}{3} + C$$

Why this is an easy question dressed as a hard one. It looks like an exponential integral and is a cubic. The only step that matters is the first, and the $(x-1)$ that cancels is why the simplification is exact rather than approximate.

Question 5

$$\int\frac{(1+\cos x)}{x+\sin x}\,dx$$

Solution. Differentiate the denominator:

$$\frac{d}{dx}\left(x+\sin x\right) = 1 + \cos x$$

the numerator again. This is the same $\int \dfrac{f’}{f}$ pattern as Question 2, disguised by having a trigonometric function in it.

Question 6

$$\int\frac{dx}{1+\cos x}$$

Solution. Use the half-angle identity $1 + \cos x = 2\cos^2\dfrac{x}{2}$:

$$\int\frac{dx}{2\cos^2\frac{x}{2}} = \frac{1}{2}\int\sec^2\frac{x}{2}\,dx = \frac{1}{2}\cdot\frac{\tan\frac{x}{2}}{1/2} + C = \tan\frac{x}{2} + C$$

Why not multiply by the conjugate. Multiplying above and below by $1-\cos x$ gives $\dfrac{1-\cos x}{\sin^2 x}$, which splits into $\operatorname{cosec}^2 x – \operatorname{cosec} x\cot x$ and integrates to $-\cot x + \operatorname{cosec} x + C$. That is the same function as $\tan\frac{x}{2}$, and it takes three times as long. Reach for the half-angle identity first whenever you see $1 \pm \cos x$ or $1 \pm \sin x$.

Question 7

$$\int\tan^2 x\,\sec^4 x\,dx$$

Solution. The presence of an even power of $\sec x$ is the signal to substitute $t = \tan x$, because then $dt = \sec^2 x\,dx$ and the remaining $\sec^2 x$ converts cleanly by $\sec^2 x = 1 + \tan^2 x$.

$$\int\tan^2 x\,\sec^2 x\cdot\sec^2 x\,dx = \int t^2\left(1+t^2\right)dt = \int\left(t^2 + t^4\right)dt = \frac{t^3}{3} + \frac{t^5}{5} + C$$

The rule worth carrying. Even power of $\sec$ → substitute $\tan x$. Odd power of $\tan$ with any $\sec$ → substitute $\sec x$. Getting this the wrong way round is the commonest way to turn a two-line integral into an impasse.

Question 8

$$\int\frac{\sin x + \cos x}{\sqrt{1+\sin 2x}}\,dx$$

Solution. The denominator is a perfect square in disguise. Since $\sin 2x = 2\sin x\cos x$,

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

so $\sqrt{1+\sin 2x} = \left|\sin x + \cos x\right|$ — and the modulus is not decoration.

Where $\sin x + \cos x > 0$ the integrand is $\dfrac{\sin x+\cos x}{\sin x+\cos x} = 1$, and the integral is $x + C$.

The domain point the key leaves out, and it is worth knowing. NCERT’s key gives “$x + c$” flat. But where $\sin x+\cos x < 0$ — for instance at $x = 3.5$ radians — the integrand is $-1$, not $+1$, and the antiderivative there is $-x + C$. Differentiating $x$ at $x=3.5$ and subtracting the integrand leaves exactly $2$, not $0$. The general answer is $\pm x + C$ according to the sign of $\sin x + \cos x$ on the interval you are working in, and an examiner asking for the indefinite integral expects $x + C$ with that interval understood.

Question 9

$$\int\sqrt{1+\sin x}\,dx$$

Solution. The same trick as Question 8. Using $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$,

$$1 + \sin x = \sin^2\frac{x}{2} + \cos^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2$$

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

Why the half-angle and not the full angle. $1+\sin x$ is a perfect square only in terms of $x/2$; there is no identity that makes it one in $x$. Questions 8 and 9 are the same idea at two different angles, which is why the exemplar puts them next to each other.

Question 10

$$\int\frac{x}{\sqrt{x}+1}\,dx \qquad \textbf{(Hint: put } \sqrt{x} = z\textbf{)}$$

Solution. With $\sqrt{x} = z$ we have $x = z^2$ and $dx = 2z\,dz$:

$$\int\frac{z^2}{z+1}\cdot 2z\,dz = 2\int\frac{z^3}{z+1}\,dz$$

Now divide, exactly as in Question 3 — $z^3 = (z+1)\left(z^2-z+1\right) – 1$:

$$2\int\left(z^2 – z + 1 – \frac{1}{z+1}\right)dz = 2\left[\frac{z^3}{3} – \frac{z^2}{2} + z – \log|z+1|\right] + C$$

Put $z = \sqrt{x}$ back, so $z^2 = x$ and $z^3 = x\sqrt{x}$:

Why the hint is given. Without it, the surd in the denominator has no obvious handle. The substitution’s whole purpose is to turn $\sqrt{x}$ into a polynomial variable so the division of Question 3 becomes available again.

Question 11

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

Solution. Rationalise by multiplying numerator and denominator inside the root by $a+x$:

$$\sqrt{\frac{a+x}{a-x}} = \sqrt{\frac{(a+x)^2}{(a-x)(a+x)}} = \frac{a+x}{\sqrt{a^2-x^2}}$$

Now split it, because one piece is a standard form and the other is an $\int f’/\sqrt{f}$:

$$\int\frac{a}{\sqrt{a^2-x^2}}\,dx + \int\frac{x}{\sqrt{a^2-x^2}}\,dx = a\sin^{-1}\frac{x}{a} – \sqrt{a^2-x^2} + C$$

The second integral comes out by putting $u = a^2-x^2$, $du = -2x\,dx$.

Why this looks different from the printed answer, and why both are right. NCERT’s key gives

$$-a\left[\cos^{-1}\frac{x}{a} + \sqrt{1-\frac{x^2}{a^2}}\right] + C$$

Expand it: $-a\sqrt{1-x^2/a^2} = -\sqrt{a^2-x^2}$, and since $\cos^{-1}\frac{x}{a} = \frac{\pi}{2} – \sin^{-1}\frac{x}{a}$,

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

So the key’s answer is ours minus the constant $\dfrac{a\pi}{2}$ — and a constant is precisely what $C$ absorbs. Two antiderivatives of the same function always differ by a constant, and an answer in $\cos^{-1}$ rather than $\sin^{-1}$ is a common way for that to show up. Do not “correct” your working because the printed form looks different; differentiate both and see.

Question 12

$$\int\frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}}\,dx \qquad \textbf{(Hint: put } x = z^4\textbf{)}$$

Solution. The hint’s exponent is chosen to clear both fractional powers at once: $4$ is the lowest common denominator of $\frac12$ and $\frac34$. With $x = z^4$, $dx = 4z^3\,dz$, $x^{1/2} = z^2$ and $x^{3/4} = z^3$:

$$\int\frac{z^2}{1+z^3}\cdot 4z^3\,dz = 4\int\frac{z^5}{1+z^3}\,dz$$

Divide — and this is the step to be careful with, because $z^5$ divided by $z^3+1$ is $z^2$ with remainder $-z^2$:

$$\frac{z^5}{z^3+1} = z^2 – \frac{z^2}{z^3+1}$$

$$4\int\left(z^2 – \frac{z^2}{z^3+1}\right)dz = 4\left[\frac{z^3}{3} – \frac{1}{3}\log\left|z^3+1\right|\right] + C$$

The second integral is $\int f’/f$ once more, since $\frac{d}{dz}(z^3+1) = 3z^2$. Putting $z = x^{1/4}$ back:

The mistake to avoid, because it is easy to make. Dividing $z^5$ by $z^3+1$ and writing the quotient as $z^2 – z + 1 – \dots$, by analogy with Question 10, gives an answer that differentiates to something else entirely. The divisor here is $z^3+1$, not $z+1$: the quotient is a single term, not three.

Question 13

$$\int\frac{\sqrt{1+x^2}}{x^4}\,dx$$

Solution. Take $x^2$ out of the root so the integrand becomes a function of $1/x^2$, which is what makes a substitution available:

$$\frac{\sqrt{1+x^2}}{x^4} = \frac{x\sqrt{1+\frac{1}{x^2}}}{x^4} = \frac{1}{x^3}\sqrt{1+\frac{1}{x^2}}$$

Put $t = 1 + \dfrac{1}{x^2}$, so $dt = -\dfrac{2}{x^3}dx$, i.e. $\dfrac{dx}{x^3} = -\dfrac{dt}{2}$:

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

Why pull out $x$ and not attempt a trigonometric substitution. $x = \tan\theta$ does work, but it lands you on $\int\cot^3\theta\operatorname{cosec}\theta\,d\theta$, which is longer. Whenever the integrand is a root of $1+x^2$ over a power of $x$, dividing through by that power first is faster.

Question 14

$$\int\frac{dx}{\sqrt{16-9x^2}}$$

Solution. Force it into the standard form $\displaystyle\int\frac{dx}{\sqrt{a^2-u^2}} = \sin^{-1}\frac{u}{a} + C$ by taking the coefficient of $x^2$ outside:

$$\int\frac{dx}{\sqrt{9\left(\frac{16}{9}-x^2\right)}} = \frac{1}{3}\int\frac{dx}{\sqrt{\left(\frac{4}{3}\right)^2-x^2}} = \frac{1}{3}\sin^{-1}\frac{x}{4/3} + C$$

The factor that gets dropped. The $\frac13$ in front comes from $\sqrt{9} = 3$ and is the part students lose. Check it by differentiating: $\frac{d}{dx}\left[\frac13\sin^{-1}\frac{3x}{4}\right] = \frac13\cdot\frac{3/4}{\sqrt{1-9x^2/16}} = \frac{1}{\sqrt{16-9x^2}}$.

Question 15

$$\int\frac{dt}{\sqrt{3t-2t^2}}$$

Solution. There is no $a^2 – u^2$ visible, so make one by completing the square. Take the $-2$ out first:

$$3t – 2t^2 = -2\left(t^2 – \frac{3t}{2}\right) = -2\left[\left(t-\frac{3}{4}\right)^2 – \frac{9}{16}\right] = 2\left[\frac{9}{16} – \left(t-\frac{3}{4}\right)^2\right]$$

$$\int\frac{dt}{\sqrt{2}\sqrt{\left(\frac{3}{4}\right)^2-\left(t-\frac{3}{4}\right)^2}} = \frac{1}{\sqrt{2}}\sin^{-1}\!\left(\frac{t-\frac34}{\frac34}\right) + C$$

Why the coefficient must come out before completing the square. Completing the square on $3t-2t^2$ with the $-2$ still attached is where the arithmetic goes wrong. Factor first, complete the square on a monic quadratic, then put the factor back as $\frac{1}{\sqrt{2}}$.

Question 16

$$\int\frac{3x-1}{\sqrt{x^2+9}}\,dx$$

Solution. This is the question in the band most worth studying, because the numerator is almost the derivative of what is under the root. $\frac{d}{dx}\left(x^2+9\right) = 2x$, and the numerator has a $3x$ in it. So split the fraction into the part that fits and the part that does not:

$$\int\frac{3x}{\sqrt{x^2+9}}\,dx – \int\frac{dx}{\sqrt{x^2+9}}$$

The first: put $u = x^2+9$, $du = 2x\,dx$, giving $\frac{3}{2}\int u^{-1/2}du = 3\sqrt{x^2+9}$. The second is the standard form $\int\frac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right| + C$.

The move to remember. A linear numerator over a root of a quadratic always splits this way: write the numerator as $\lambda\cdot(\text{derivative of the quadratic}) + \mu$, and the integral becomes a root plus a standard form. It works for every question of this shape and there are several later in the unit.

Question 17

$$\int\sqrt{5-2x+x^2}\,dx$$

Solution. Complete the square under the root — with the $x^2$ already monic, this is direct:

$$x^2 – 2x + 5 = (x-1)^2 + 4$$

so with $u = x-1$ and $a = 2$ this is $\int\sqrt{u^2+a^2}\,du$, whose standard result is

$$\int\sqrt{u^2+a^2}\,du = \frac{u}{2}\sqrt{u^2+a^2} + \frac{a^2}{2}\log\left|u+\sqrt{u^2+a^2}\right| + C$$

With $a^2 = 4$, the second coefficient is $\frac{4}{2} = 2$:

Why rewrite the terms in $x$ at the end. Leaving the answer in $u$ is incomplete, and leaving $\sqrt{u^2+4}$ unexpanded hides that it equals $\sqrt{x^2-2x+5}$ — the same expression the question started from. Writing it back makes the answer checkable by inspection.

Common mistakes

• Treating Questions 1 and 2 as evaluations. They say verify, so the answer is given and the marks are for the demonstration. Integrating the left side is the better demonstration; differentiating the right side is also valid but leaves you having to argue the two forms agree.
• Dividing wrongly in Question 12. $z^5 \div (z^3+1)$ is $z^2$ with remainder $-z^2$ — a single term, not the three-term quotient that $z^3 \div (z+1)$ gives in Question 10. This is the one error on the page that produces a confidently wrong answer rather than a dead end.
• Dropping the modulus in Question 8. $\sqrt{1+\sin 2x} = \left|\sin x+\cos x\right|$, and where that is negative the integral is $-x$, not $x$. The printed key’s bare “$x+c$” carries an unstated interval.
• Assuming a different-looking printed answer means yours is wrong, as in Question 11. Two antiderivatives of the same function differ by a constant; the key’s $\cos^{-1}$ form and the $\sin^{-1}$ form here differ by exactly $\frac{a\pi}{2}$.
• Losing the constant in front in Question 14. Pulling $9$ out of the root leaves a $\frac13$ outside, and that factor is dropped more often than any other step in the band.
• Completing the square before factoring out the coefficient in Question 15. On $3t-2t^2$ the $-2$ has to come out first; doing it in the other order is where the $\frac34$ and the $\frac{9}{16}$ go wrong.
• Substituting $\sec x$ where $\tan x$ is wanted in Question 7. The even power of $\sec$ is the signal for $t=\tan x$. The other substitution leaves an odd power of $\sec$ with nothing to pair it with.

Practise next

• Exemplar Short Answer Q18–34, the second half of this band, where the splitting move from Question 16 and the completing-the-square of Questions 15 and 17 come back on harder quadratics.
• Exemplar Long Answer Q35–47, which is where definite integrals and the properties of $\int_0^a$ appear.
• NCERT textbook Exercise 7.1 and Exercise 7.2 for the standard forms and substitutions these questions assume you already have.
• NCERT textbook Exercise 7.4, the textbook’s own set on $\frac{1}{\sqrt{a^2-x^2}}$ and its relatives — Questions 14 to 17 here are the exemplar’s harder versions of it.