NCERT Exemplar Solutions Class 12 Maths Integrals — Objective and Fill in the Blanks Q48–63

NCERT Exemplar Problems, Class 12 Mathematics — Integrals, Objective Type questions 48 to 58 and Fill in the Blanks 59 to 63. All 16 questions solved.

The last band of the unit is eleven multiple-choice questions and five fill-in-the-blanks, and almost every one of them is a single recognition away from being immediate. Question 48 is two lines once you see the difference of squares; Question 51 is two lines once you see $f+f’$; Question 63 needs no integration at all. The band is not testing stamina — it is testing whether you can see the structure before you start.

Two things are worth knowing before you begin. The book’s own numbering instructions are wrong twice here. The line above Question 48 reads “Choose the correct option from given four options in each of the Exercises from 48 to 63”, but the multiple-choice questions stop at 58; and the “Fill in the blanks” instruction is printed below Question 60 and says “Exercise 60 to 63”, when Question 59 is a fill-in too. Both are typographic slips, and neither changes what any question asks.

Key insight. In an objective question the options are data. Three of the four in Questions 53 and 56 can be eliminated by differentiating the given form and matching a single coefficient — far faster than integrating from scratch. And where the options differ only by a sign or by a squared denominator, as in Question 51, differentiating each candidate is not a fallback, it is the method.

Question 48

$$\int\frac{\cos 2x-\cos 2\theta}{\cos x-\cos\theta}\,dx \text{ is equal to}$$

Solution. Use the same form of the double angle on both terms, $\cos 2A = 2\cos^2A-1$, so the two $-1$s cancel:

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

The denominator cancels outright, leaving $\int 2(\cos x+\cos\theta)\,dx$. Remember that $\theta$ is a constant here, so $\cos\theta$ integrates to $x\cos\theta$.

Why the $-1$ form and not $1-2\sin^2$. Either works, but $2\cos^2A-1$ produces a difference of squares in $\cos$, which is what the denominator is written in. Matching the identity to the denominator is the whole trick.

Question 49

$$\int\frac{dx}{\sin(x-a)\sin(x-b)} \text{ is equal to}$$

Solution. The standard device is to manufacture a $1$ out of the constant angle. Since $(x-a)-(x-b) = b-a$,

$$\sin(b-a) = \sin\left[(x-a)-(x-b)\right] = \sin(x-a)\cos(x-b)-\cos(x-a)\sin(x-b)$$

Divide throughout by $\sin(x-a)\sin(x-b)$ and the right-hand side becomes $\cot(x-b)-\cot(x-a)$. Hence

$$\frac{1}{\sin(x-a)\sin(x-b)} = \operatorname{cosec}(b-a)\left[\cot(x-b)-\cot(x-a)\right]$$

and each cotangent integrates to a logarithm of a sine.

Two options can be discarded without any work. $\sin(b-a)$ is a multiplier in (A) and (D), but the constant enters as a divisor — a smaller angle difference must make the integral larger, not smaller. That leaves (B) and (C), which differ only in which sine is on top, and the expansion above settles it.

Question 50

$$\int\tan^{-1}\sqrt{x}\,dx \text{ is equal to}$$

Solution. Integrate by parts against $1$, which is the only available choice:

$$x\tan^{-1}\sqrt x – \int x\cdot\frac{1}{1+x}\cdot\frac{1}{2\sqrt x}\,dx = x\tan^{-1}\sqrt x – \frac12\int\frac{\sqrt x}{1+x}\,dx$$

For the remaining integral put $x=t^2$, $dx=2t\,dt$:

$$\frac12\int\frac{t}{1+t^2}\cdot 2t\,dt = \int\frac{t^2}{1+t^2}\,dt = \int\left(1-\frac{1}{1+t^2}\right)dt = t-\tan^{-1}t$$

which is $\sqrt x – \tan^{-1}\sqrt x$. Subtracting gives $x\tan^{-1}\sqrt x – \sqrt x + \tan^{-1}\sqrt x$.

Read the question from the page, not from a summary. This is $\tan^{-1}\sqrt{x}$, not $\tan^{-1}x$ — and for $\tan^{-1}x$ the answer would be the familiar $x\tan^{-1}x-\frac12\log(1+x^2)$, which is not among the options at all. Option (B) is exactly the trap for anyone who drops the root.

Question 51

$$\int e^{x}\left(\frac{1-x}{1+x^2}\right)^{2}dx \text{ is equal to}$$

Solution. Expand the square and split it so that the $f+f’$ pattern appears:

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

Take $f(x)=\dfrac{1}{1+x^2}$. Then $f'(x) = -\dfrac{2x}{\left(1+x^2\right)^{2}}$, and the bracket above is exactly $f+f’$. So $\int e^{x}\left[f(x)+f'(x)\right]dx = e^{x}f(x)+C$.

The printed answer key is wrong here — it gives (C). This is worth stating plainly rather than working around, because a student checking their work against the back of the book will conclude they have made a mistake when they have not. Differentiating (C) gives

$$\frac{d}{dx}\frac{e^{x}}{\left(1+x^2\right)^{2}} = \frac{e^{x}\left(1+x^2\right)-4x\,e^{x}}{\left(1+x^2\right)^{3}}$$

which is not the integrand — the denominator is a cube. Differentiating (A) gives $\dfrac{e^{x}\left(1+x^2-2x\right)}{\left(1+x^2\right)^{2}} = \dfrac{e^{x}(1-x)^2}{\left(1+x^2\right)^{2}}$, which is the integrand exactly. (A) is the answer.

Question 52

$$\int\frac{x^{9}}{\left(4x^{2}+1\right)^{6}}\,dx \text{ is equal to}$$

Solution. A sixth power in the denominator rules out expansion, so force the substitution instead. Take $x^{12}$ out of the denominator — twelve powers, because $\left(4x^2+1\right)^6 = x^{12}\left(4+\frac{1}{x^2}\right)^{6}$:

$$\int\frac{x^{9}\,dx}{x^{12}\left(4+\frac{1}{x^{2}}\right)^{6}} = \int\frac{dx}{x^{3}\left(4+\frac{1}{x^{2}}\right)^{6}}$$

Now $u = 4+\dfrac{1}{x^2}$ gives $du = -\dfrac{2}{x^{3}}dx$, so $\dfrac{dx}{x^3} = -\dfrac{du}{2}$:

$$-\frac12\int u^{-6}\,du = -\frac12\cdot\frac{u^{-5}}{-5} = \frac{u^{-5}}{10}$$

Option (C) is not a function of $x$ at all in its bracket — $(1+4)^{-5}$ is a constant. Scanning the options for one that could not possibly be an antiderivative is often quicker than integrating, and it is free.

Question 53

$$\text{If } \int\frac{dx}{(x+2)\left(x^{2}+1\right)} = a\log\left|1+x^{2}\right| + b\tan^{-1}x + \frac15\log|x+2| + C, \text{ then}$$

Solution. Decompose with a linear numerator over the irreducible quadratic:

$$\frac{1}{(x+2)\left(x^2+1\right)} = \frac{A}{x+2} + \frac{Bx+C_1}{x^2+1}$$

Putting $x=-2$ gives $A=\frac15$ — which the question has already told us by printing $\frac15\log|x+2|$, a useful confirmation that the decomposition is set up right. Comparing coefficients of $x^2$ gives $A+B=0$, so $B=-\frac15$; comparing constants gives $A+2C_1=1$, so $C_1=\frac25$.

The $Bx$ term integrates to $\frac{B}{2}\log\left(1+x^2\right)$, so $a = \frac{B}{2} = -\frac{1}{10}$, and $C_1$ is the coefficient of the arctangent, so $b=\frac25$.

The factor of $\frac12$ is where this question is lost. $\int\frac{x\,dx}{1+x^2} = \frac12\log\left(1+x^2\right)$, so $a$ is half the coefficient $B$, not $B$ itself. Option (A) is what you get by taking $b$’s sign from $B$ instead of from $C_1$; option (D) is what you get by dropping both.

Question 54

$$\int\frac{x^{3}}{x+1}\,dx \text{ is equal to}$$

Solution. The numerator has higher degree than the denominator, so divide first. Dividing $x^3$ by $x+1$:

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

Integrating term by term gives $\frac{x^3}{3}-\frac{x^2}{2}+x-\log|x+1|$.

Two of the options have $\log|1-x|$, which no integrand with $x+1$ in the denominator can produce. Checking the logarithm alone eliminates half the list before the polynomial is even divided.

Question 55

$$\int\frac{x+\sin x}{1+\cos x}\,dx \text{ is equal to}$$

Solution. Halve every angle. With $1+\cos x = 2\cos^2\frac x2$ and $\sin x = 2\sin\frac x2\cos\frac x2$, the integrand splits into

$$\frac{x}{2}\sec^2\frac{x}{2} + \tan\frac{x}{2}$$

Integrate the first piece by parts, taking $x$ as the first function, since $\int\frac12\sec^2\frac x2\,dx = \tan\frac x2$:

$$x\tan\frac{x}{2} – \int\tan\frac{x}{2}\,dx$$

and the second piece of the integrand is exactly the $\int\tan\frac x2\,dx$ that parts has just subtracted. They cancel.

This is the $\int e^{x}\left[f+f’\right]$ idea without the exponential. An integrand that is a function plus something whose parts-integral reproduces that function collapses to a single product. Option (B) is the trap for reading $\frac{f’}{f}$ into the fraction — but the numerator is not the derivative of the denominator, and it is worth checking rather than assuming.

Question 56

$$\text{If } \int\frac{x^{3}\,dx}{\sqrt{1+x^{2}}} = a\left(1+x^{2}\right)^{3/2} + b\sqrt{1+x^{2}} + C, \text{ then}$$

Solution. Split the odd power as $x^3 = x^2\cdot x$, keeping one $x$ back for the substitution $u=1+x^2$, $x\,dx=\frac{du}{2}$, $x^2=u-1$:

$$\int\frac{(u-1)}{\sqrt u}\cdot\frac{du}{2} = \frac12\int\left(u^{1/2}-u^{-1/2}\right)du = \frac12\left[\frac23u^{3/2}-2u^{1/2}\right] = \frac13u^{3/2}-u^{1/2}$$

Check the sign without integrating. For large $x$ the integrand behaves like $x^2$, so the antiderivative must grow like $\frac{x^3}{3}$ — which fixes $a=+\frac13$ immediately and eliminates (B) and (C). Only the sign of $b$ then remains, and $u^{-1/2}$ enters the bracket with a minus.

Question 57

$$\int_{-\pi/4}^{\pi/4}\frac{dx}{1+\cos 2x} \text{ is equal to}$$

Solution. $1+\cos 2x = 2\cos^2x$, so the integrand is $\frac12\sec^2x$ and the antiderivative is $\frac12\tan x$:

$$\frac12\left[\tan x\right]_{-\pi/4}^{\pi/4} = \frac12\left(1-(-1)\right) = 1$$

The integrand is even, so the interval could have been halved to $2\times\frac12\left[\tan x\right]_0^{\pi/4}$. Either route is one line here, but on a harder integrand the evenness is what makes the limits manageable, and it is worth checking for by habit.

Question 58

$$\int_{0}^{\pi/2}\sqrt{1-\sin 2x}\;dx \text{ is equal to}$$

Solution. Since $\sin 2x = 2\sin x\cos x$ and $\sin^2x+\cos^2x=1$,

$$1-\sin 2x = \left(\sin x-\cos x\right)^{2} \qquad\Longrightarrow\qquad \sqrt{1-\sin 2x} = \left|\sin x-\cos x\right|$$

The modulus is the whole question. On $\left[0,\frac\pi4\right)$ cosine is the larger, and on $\left(\frac\pi4,\frac\pi2\right]$ sine is, so the interval must be split at $\frac\pi4$:

$$\int_0^{\pi/4}\left(\cos x-\sin x\right)dx = \left[\sin x+\cos x\right]_0^{\pi/4} = \sqrt2-1$$

$$\int_{\pi/4}^{\pi/2}\left(\sin x-\cos x\right)dx = \left[-\cos x-\sin x\right]_{\pi/4}^{\pi/2} = -1+\sqrt2$$

The two halves are equal, which they must be — the integrand is symmetric about $x=\frac\pi4$.

Dropping the modulus gives zero, and zero is not an option — which is itself the hint. Any square root of a perfect square over an interval where the base changes sign has to be split at the crossing point.

Question 59

$$\int_{0}^{\pi/2}\cos x\,e^{\sin x}\,dx \text{ is equal to } \rule{2cm}{0.4pt}$$

Solution. The derivative of $\sin x$ is sitting in front of the exponential, so $u=\sin x$ makes this immediate, with $u$ running from $0$ to $1$:

$$\int_0^1 e^{u}\,du = \left[e^{u}\right]_0^{1}$$

Question 60

$$\int\frac{x+3}{(x+4)^{2}}\,e^{x}\,dx = \rule{2cm}{0.4pt}$$

Solution. Write the numerator in terms of the denominator’s base — $x+3 = (x+4)-1$ — and split:

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

With $f(x)=\dfrac{1}{x+4}$ we have $f'(x) = -\dfrac{1}{(x+4)^2}$, so the bracket is $f+f’$ and $\int e^{x}\left[f+f’\right]dx = e^{x}f(x)$.

The same pattern as Questions 51 and 55, three times in one band. Whenever $e^x$ multiplies a rational function, rewrite the numerator in terms of the denominator before doing anything else — it is the standard way the $f+f’$ structure is hidden.

Question 61

$$\text{If } \int_{0}^{a}\frac{1}{1+4x^{2}}\,dx = \frac{\pi}{8}, \text{ then } a = \rule{2cm}{0.4pt}$$

Solution. Write $1+4x^2$ as $4\left(x^2+\frac14\right)$, or simply substitute $u=2x$. Either way

$$\int_0^a\frac{dx}{1+4x^2} = \frac12\left[\tan^{-1}2x\right]_0^{a} = \frac12\tan^{-1}2a$$

Setting that equal to $\frac\pi8$ gives $\tan^{-1}2a = \frac\pi4$, so $2a=1$.

The factor of $\frac12$ in front is the point. $\int\frac{dx}{1+4x^2}$ is not $\tan^{-1}2x$; the chain rule leaves a $\frac12$ behind, and forgetting it produces $\tan^{-1}2a=\frac\pi8$ and the wrong value $a=\frac12\tan\frac\pi8$.

Question 62

$$\int\frac{\sin x}{3+4\cos^{2}x}\,dx = \rule{2cm}{0.4pt}$$

Solution. The numerator is $-\,\frac{d}{dx}\cos x$ and the denominator is a function of $\cos x$ alone, so $u=\cos x$, $du=-\sin x\,dx$:

$$-\int\frac{du}{3+4u^{2}} = -\frac14\int\frac{du}{\frac34+u^{2}} = -\frac14\cdot\frac{2}{\sqrt3}\tan^{-1}\frac{2u}{\sqrt3}$$

using $\int\frac{du}{u^2+k^2} = \frac1k\tan^{-1}\frac uk$ with $k=\frac{\sqrt3}{2}$.

Expect the minus sign. As $x$ increases through $(0,\pi)$ the integrand is positive, and $\cos x$ is falling — so an antiderivative written in $\cos x$ must decrease as its argument increases. A sign that looks wrong in a $u$-substitution usually is not; here it is exactly right.

Question 63

$$\text{The value of } \int_{-\pi}^{\pi}\sin^{3}x\,\cos^{2}x\,dx \text{ is } \rule{2cm}{0.4pt}$$

Solution. No integration is required. $\sin^3x$ is odd and $\cos^2x$ is even, so their product is odd, and the interval $[-\pi,\pi]$ is symmetric about the origin. For any odd $f$, $\int_{-a}^{a}f(x)\,dx = 0$.

The unit ends where it should. Parity is the cheapest property in the chapter and the last question is a pure test of whether you check for it before reaching for a substitution. Working this one out by expanding $\sin^3x = \frac{3\sin x-\sin 3x}{4}$ takes five minutes and arrives at the same zero.

Common mistakes

• Reading Question 50 as $\tan^{-1}x$. It is $\tan^{-1}\sqrt x$, and the two have completely different antiderivatives. Option (B) is placed to catch exactly this.
• Trusting the printed key on Question 51. The book gives (C); the answer is (A), and differentiating either option settles it in one line. When a key disagrees with a derivative you can check yourself, the derivative wins.
• Halving nothing in Question 53. $a$ is $\frac{B}{2}$, not $B$, because $\int\frac{x\,dx}{1+x^2}$ carries a factor of $\frac12$. The same $\frac12$ is what Question 61 is built on.
• Dropping the modulus in Question 58. $\sqrt{(\sin x-\cos x)^2}$ is $\left|\sin x-\cos x\right|$, and on $\left[0,\frac\pi2\right]$ that expression changes sign at $\frac\pi4$. Without the split the answer comes out as $0$.
• Integrating Question 63. It is odd over a symmetric interval, so the value is zero by inspection. Any symmetric limits should prompt a parity check before anything else.
• Missing $f+f’$ in Questions 51, 55 and 60. All three collapse to a single product once the numerator is rewritten in terms of the denominator. Expanding and integrating term by term does not work in any of them, because the individual pieces are not separately elementary.
• Treating $\theta$ as a variable in Question 48. It is a constant, so $\int\cos\theta\,dx = x\cos\theta$. Options (C) and (D) both carry a spurious factor of $2$ that comes from mishandling this.

Practise next

• Exemplar Long Answer Q35–47, where the partial fractions of Questions 53 and 54 and the $f+f’$ pattern of Question 51 are set as full-length problems.
• Exemplar Short Answer Q1–17, the start of the unit, if any of the standard forms used above were unfamiliar.
• NCERT textbook Exercise 7.6 for integration by parts, which Questions 50, 55 and the whole $e^{x}\left[f+f’\right]$ family rest on.
• NCERT textbook Exercise 7.10 for the parity and symmetry properties that decide Questions 57, 58 and 63 without any integration.