NCERT Exemplar Solutions Class 12 Maths Integrals — Long Answer Q35–47

NCERT Exemplar Problems, Class 12 Mathematics — Integrals, Long Answer questions 35 to 47. All 13 questions solved.

Thirteen questions, and the jump from the Short Answer band is real: these are the integrals that need a plan before a pen. Four run on partial fractions where the fraction must first be seen as a fraction in $x^2$ rather than in $x$; three are definite integrals that collapse the moment the right property is applied and are close to impossible without it; and two — Questions 43 and 44 — are the standard hard cases that reward a substitution nobody guesses twice.

The techniques the band actually runs on are few: partial fractions, integration by parts in the form $\int e^{x}\left[f(x)+f'(x)\right]dx = e^{x}f(x)$, the reduction of $\int e^{ax}\cos bx\,dx$ to a standard closed form, and the two properties $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ and $\int_{-a}^{a}$ of an even function.

Key insight. Nine of these thirteen are decided before any integration happens — by factorising the denominator into quadratics rather than linears (35, 36), by recognising $f+f’$ under an exponential (39), by halving the angle (41), or by reflecting the interval (37, 46, 47). The integration that follows is routine in every case. What the band tests is whether you can see which of the four it is.

Question 35

$$\int\frac{x^2\,dx}{x^4-x^2-12}$$

Solution. The denominator is a quadratic in $x^2$, so factorise it as one: $x^4-x^2-12=(x^2-4)(x^2+3)$. Splitting into partial fractions in $x^2$ — not in $x$ — gives

$$\frac{x^2}{(x^2-4)(x^2+3)}=\frac{A}{x^2-4}+\frac{B}{x^2+3},\qquad x^2 = A(x^2+3)+B(x^2-4)$$

Putting $x^2=4$ gives $4=7A$, so $A=\frac47$; putting $x^2=-3$ gives $-3=-7B$, so $B=\frac37$. The two pieces are now standard forms — the first a difference of squares, the second an arctangent:

$$\frac47\cdot\frac{1}{4}\log\left|\frac{x-2}{x+2}\right| + \frac37\cdot\frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}$$

Why not factor all the way down. $x^2-4$ splits further into $(x-2)(x+2)$, and doing so turns a two-term decomposition into a three-term one for no gain — $x^2+3$ is irreducible regardless, so the arctangent is coming either way. Stopping at quadratics is the whole economy of the method here.

Question 36

$$\int\frac{x^2\,dx}{\left(x^2+a^2\right)\left(x^2+b^2\right)}$$

Solution. Same structure as Question 35 and the same move: treat $x^2$ as the variable.

$$x^2 = A\left(x^2+b^2\right)+B\left(x^2+a^2\right)$$

Putting $x^2=-a^2$ gives $-a^2 = A\left(b^2-a^2\right)$, so $A=\dfrac{a^2}{a^2-b^2}$; putting $x^2=-b^2$ gives $B=\dfrac{b^2}{b^2-a^2}$. Each term is now an arctangent, and the $a^2$ on top cancels one factor of $a$ from the $\frac1a$ that $\int\frac{dx}{x^2+a^2}$ contributes:

$$\frac{a^2}{a^2-b^2}\cdot\frac1a\tan^{-1}\frac{x}{a} + \frac{b^2}{b^2-a^2}\cdot\frac1b\tan^{-1}\frac{x}{b}$$

The answer is undefined when $a^2=b^2$, and rightly so — the integral is then $\int\frac{x^2dx}{(x^2+a^2)^2}$, a genuinely different problem with no arctangent-difference form. A parameter in the denominator of an answer is always worth reading as a statement about which cases the method covers.

Question 37

$$\int_0^{\pi}\frac{x}{1+\sin x}\,dx$$

Solution. The $x$ on top blocks every substitution, which is the signal to reflect the interval instead. Using $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ with $a=\pi$, and $\sin(\pi-x)=\sin x$:

$$I = \int_0^{\pi}\frac{\pi-x}{1+\sin x}\,dx \qquad\Longrightarrow\qquad 2I = \pi\int_0^{\pi}\frac{dx}{1+\sin x}$$

The $x$ is gone. For what remains, the half-angle substitution $t=\tan\frac{x}{2}$ turns $1+\sin x$ into a perfect square:

$$1+\sin x = 1+\frac{2t}{1+t^2} = \frac{(1+t)^2}{1+t^2},\qquad dx=\frac{2\,dt}{1+t^2}$$

so that $\displaystyle\int_0^{\pi}\frac{dx}{1+\sin x}=\int_0^{\infty}\frac{2\,dt}{(1+t)^2}=\left[\frac{-2}{1+t}\right]_0^{\infty}=2$. Hence $2I=2\pi$.

The reflection is doing all the work. Note that it succeeds because $\sin x$ is symmetric about $x=\frac{\pi}{2}$ over exactly this interval. The same trick on $\int_0^{\pi/2}$ would not remove the $x$, because $\sin\!\left(\frac\pi2-x\right)=\cos x$ is a different function.

Question 38

$$\int\frac{2x-1}{(x-1)(x+2)(x-3)}\,dx$$

Solution. Three distinct linear factors, so three constants, each read off by covering its own factor and substituting that root:

$$\frac{2x-1}{(x-1)(x+2)(x-3)} = \frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}$$

At $x=1$: $A=\dfrac{1}{(3)(-2)}=-\dfrac16$. At $x=-2$: $B=\dfrac{-5}{(-3)(-5)}=-\dfrac13$. At $x=3$: $C=\dfrac{5}{(2)(5)}=\dfrac12$.

The book prints this as a single logarithm, $\log\left|\dfrac{\sqrt{x-3}}{(x-1)^{1/6}(x+2)^{1/3}}\right|+c$, which is the same thing with the coefficients pushed inside as fractional powers. Either is correct; three separate terms are easier to check and easier to differentiate back.

Question 39

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

Solution. Substitute $t=\tan^{-1}x$, so $x=\tan t$ and $dx=\sec^2 t\,dt=(1+x^2)\,dt$. The $1+x^2$ in the denominator cancels exactly, leaving

$$\int e^{t}\left(1+\tan t+\tan^2 t\right)dt = \int e^{t}\left(\sec^2 t + \tan t\right)dt$$

using $1+\tan^2 t=\sec^2 t$. Now read the bracket as $f(t)+f'(t)$ with $f(t)=\tan t$, since $\frac{d}{dt}\tan t=\sec^2 t$. That is precisely the pattern $\int e^{t}\left[f(t)+f'(t)\right]dt=e^{t}f(t)+C$:

$$e^{t}\tan t + C$$

The $+x$ in the numerator is the entire question. Without it the integral is simply $e^{\tan^{-1}x}$; with it the expression becomes $f+f’$ and the answer picks up a factor of $x$. Splitting the fraction into $1+\frac{x}{1+x^2}$ and integrating term by term does not work, because the second piece is not integrable against $e^{\tan^{-1}x}$ on its own.

Question 40

$$\int\sin^{-1}\sqrt{\frac{x}{a+x}}\,dx \qquad \textbf{(Hint: Put } x = a\tan^2\theta\textbf{)}$$

Solution. The hint is aimed at the inverse sine, not at the root. With $x=a\tan^2\theta$,

$$\frac{x}{a+x} = \frac{a\tan^2\theta}{a\sec^2\theta} = \sin^2\theta$$

so $\sin^{-1}\sqrt{\frac{x}{a+x}}=\theta$ outright, and $dx = 2a\tan\theta\sec^2\theta\,d\theta$. The integral becomes

$$2a\int\theta\,\tan\theta\sec^2\theta\,d\theta$$

Integrate by parts with $\theta$ as the first function, noting $\int\tan\theta\sec^2\theta\,d\theta=\frac{\tan^2\theta}{2}$:

$$2a\left[\theta\cdot\frac{\tan^2\theta}{2}-\int\frac{\tan^2\theta}{2}\,d\theta\right] = a\,\theta\tan^2\theta – a\int\left(\sec^2\theta-1\right)d\theta = a\,\theta\tan^2\theta – a\tan\theta + a\theta$$

Now $\tan\theta=\sqrt{\frac{x}{a}}$, $\tan^2\theta=\frac{x}{a}$ and $\theta=\tan^{-1}\sqrt{\frac{x}{a}}$, so $a\theta\tan^2\theta = x\theta$ and $a\tan\theta=\sqrt{ax}$:

Why $\tan^{-1}$ appears in an answer that started with $\sin^{-1}$. They are the same angle: $\theta=\sin^{-1}\sqrt{\frac{x}{a+x}}=\tan^{-1}\sqrt{\frac{x}{a}}$, since a right triangle with opposite $\sqrt{x}$ and hypotenuse $\sqrt{a+x}$ has adjacent $\sqrt{a}$. Writing the answer in $\tan^{-1}$ keeps it in the variable the substitution produced.

Question 41

$$\int_{\pi/3}^{\pi/2}\frac{\sqrt{1+\cos x}}{\left(1-\cos x\right)^{5/2}}\,dx$$

Solution. Both half-angle identities apply at once: $1+\cos x = 2\cos^2\frac{x}{2}$ and $1-\cos x = 2\sin^2\frac{x}{2}$. On $\left[\frac\pi3,\frac\pi2\right]$ the half-angle lies in $\left[\frac\pi6,\frac\pi4\right]$, where both sine and cosine are positive, so the roots come out without absolute values:

$$\frac{\sqrt2\,\cos\frac{x}{2}}{2^{5/2}\sin^5\frac{x}{2}} = \frac14\cdot\frac{\cos\frac{x}{2}}{\sin^5\frac{x}{2}}$$

Substituting $u=\sin\frac{x}{2}$, so $\cos\frac{x}{2}\,dx = 2\,du$, with $u$ running from $\sin\frac\pi6=\frac12$ to $\sin\frac\pi4=\frac{1}{\sqrt2}$:

$$\frac14\int 2u^{-5}\,du = \left[-\frac{1}{8u^4}\right]_{1/2}^{1/\sqrt2} = -\frac{1}{8\cdot\frac14}+\frac{1}{8\cdot\frac{1}{16}} = -\frac12+2$$

The fractional power is a disguise. A $\frac52$ exponent looks forbidding until the half-angle identity turns $(1-\cos x)^{5/2}$ into $2^{5/2}\sin^5\frac{x}{2}$ — an ordinary fifth power. Any integrand mixing $1+\cos x$ with $1-\cos x$ should be halved on sight.

Question 42

$$\int e^{-3x}\cos^3x\,dx$$

Solution. There is no product rule that handles $\cos^3$ against an exponential, so first linearise the cube using $\cos 3x = 4\cos^3x-3\cos x$, which rearranges to

$$\cos^3x = \frac{3\cos x+\cos 3x}{4}$$

Each piece is then the standard form $\displaystyle\int e^{ax}\cos bx\,dx = \frac{e^{ax}\left(a\cos bx + b\sin bx\right)}{a^2+b^2}$, here with $a=-3$. For $b=1$ the denominator is $10$; for $b=3$ it is $18$:

$$\frac34\cdot\frac{e^{-3x}(\sin x-3\cos x)}{10} + \frac14\cdot\frac{e^{-3x}\left(3\sin 3x-3\cos 3x\right)}{18}$$

Linearise before you integrate, always. Attacking $\int e^{-3x}\cos^3x\,dx$ by parts directly generates $\cos^2x\sin x$ terms and never closes. Reducing every power of a trigonometric function to multiple angles first is what makes the exponential case finite.

Question 43

$$\int\sqrt{\tan x}\,dx \qquad \textbf{(Hint: Put } \tan x = t^2\textbf{)}$$

Solution. With $\tan x=t^2$, differentiating gives $\sec^2x\,dx = 2t\,dt$, and since $\sec^2x=1+t^4$,

$$\int \sqrt{\tan x}\,dx = \int t\cdot\frac{2t\,dt}{1+t^4} = \int\frac{2t^2}{1+t^4}\,dt$$

Now the standard split: write $2t^2 = (t^2+1)+(t^2-1)$ and handle the halves separately, dividing each by $t^2$ top and bottom.

For $\displaystyle\int\frac{t^2+1}{t^4+1}dt=\int\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt$, put $u=t-\frac1t$, so $du=\left(1+\frac{1}{t^2}\right)dt$ and $t^2+\frac{1}{t^2}=u^2+2$:

$$\int\frac{du}{u^2+2} = \frac{1}{\sqrt2}\tan^{-1}\frac{u}{\sqrt2} = \frac{1}{\sqrt2}\tan^{-1}\left(\frac{t^2-1}{\sqrt2\,t}\right)$$

For $\displaystyle\int\frac{t^2-1}{t^4+1}dt$, put $v=t+\frac1t$, so $dv=\left(1-\frac{1}{t^2}\right)dt$ and $t^2+\frac{1}{t^2}=v^2-2$:

$$\int\frac{dv}{v^2-2} = \frac{1}{2\sqrt2}\log\left|\frac{v-\sqrt2}{v+\sqrt2}\right| = \frac{1}{2\sqrt2}\log\left|\frac{t^2-\sqrt2\,t+1}{t^2+\sqrt2\,t+1}\right|$$

Restoring $t=\sqrt{\tan x}$, so that $t^2=\tan x$ and $\sqrt2\,t=\sqrt{2\tan x}$:

This is the one to learn by heart. The $\pm$ split of $2t^2$ into $(t^2+1)+(t^2-1)$, and the paired substitutions $t\mp\frac1t$ that follow from it, solve the whole family $\int\frac{t^2\,dt}{t^4+1}$, $\int\frac{dt}{t^4+1}$ and $\int\frac{t^2\pm1}{t^4+1}dt$. It is the standard technique for a quartic denominator with no real roots, and it appears in every serious integral problem set.

Question 44

$$\int_0^{\pi/2}\frac{dx}{\left(a^2\cos^2x+b^2\sin^2x\right)^2} \qquad \textbf{(Hint: Divide numerator and denominator by }\cos^4x\textbf{)}$$

Solution. Following the hint, the squared denominator contributes $\cos^4x$, so the numerator gains $\sec^4x$:

$$\int_0^{\pi/2}\frac{\sec^4x\,dx}{\left(a^2+b^2\tan^2x\right)^2}$$

Writing $\sec^4x = \sec^2x\cdot\left(1+\tan^2x\right)$ and substituting $t=\tan x$, $dt=\sec^2x\,dx$, with $t$ running from $0$ to $\infty$:

$$\int_0^{\infty}\frac{\left(1+t^2\right)dt}{\left(a^2+b^2t^2\right)^2}$$

Now a second substitution clears the square. Put $bt = a\tan\phi$, so $t=\frac ab\tan\phi$, $dt=\frac ab\sec^2\phi\,d\phi$ and $a^2+b^2t^2=a^2\sec^2\phi$, with $\phi:0\to\frac\pi2$:

$$\frac{a}{b\,a^4}\int_0^{\pi/2}\left(1+\frac{a^2}{b^2}\tan^2\phi\right)\cos^2\phi\,d\phi = \frac{1}{a^3b}\int_0^{\pi/2}\left(\cos^2\phi+\frac{a^2}{b^2}\sin^2\phi\right)d\phi$$

Both $\int_0^{\pi/2}\cos^2$ and $\int_0^{\pi/2}\sin^2$ equal $\frac\pi4$, so the bracket contributes $\frac{\pi}{4}\left(1+\frac{a^2}{b^2}\right)=\frac{\pi\left(a^2+b^2\right)}{4b^2}$:

Check it against the case you know. Setting $a=b$ makes the integrand $\frac{1}{a^4}$ over $\left[0,\frac\pi2\right]$, which integrates to $\frac{\pi}{2a^4}$ — and the answer gives $\frac{\pi}{4}\cdot\frac{2a^2}{a^6}=\frac{\pi}{2a^4}$. A special case that can be computed by hand is the cheapest possible check on a two-parameter answer, and it should be habit.

Question 45

$$\int_0^{1}x\log(1+2x)\,dx$$

Solution. By parts with $\log(1+2x)$ as the first function — it differentiates to something rational, which is the whole point of choosing it:

$$\left[\frac{x^2}{2}\log(1+2x)\right]_0^{1} – \int_0^{1}\frac{x^2}{2}\cdot\frac{2}{1+2x}\,dx = \frac{\log 3}{2} – \int_0^1\frac{x^2}{1+2x}\,dx$$

The remaining integrand is improper as a fraction, so divide out first. Since $x^2=(1+2x)\left(\frac x2-\frac14\right)+\frac14$,

$$\int_0^1\frac{x^2\,dx}{1+2x} = \int_0^1\left(\frac x2-\frac14\right)dx + \frac14\int_0^1\frac{dx}{1+2x} = \left[\frac{x^2}{4}-\frac x4\right]_0^1 + \frac18\log 3 = \frac{\log 3}{8}$$

The polynomial part vanishes exactly at the upper limit, leaving $\frac{\log 3}{2}-\frac{\log 3}{8}$.

Divide before you integrate a rational function. Feeding $\frac{x^2}{1+2x}$ straight to partial fractions is impossible — the degree of the numerator is not smaller than that of the denominator, so long division is not a shortcut but a prerequisite.

Question 46

$$\int_0^{\pi}x\log\sin x\,dx$$

Solution. Reflect the interval, exactly as in Question 37: $\sin(\pi-x)=\sin x$, so

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

Since $\sin x$ is symmetric about $x=\frac\pi2$, the remaining integral is $2\int_0^{\pi/2}\log\sin x\,dx$. Call that inner integral $J$. Because $\int_0^{\pi/2}\log\sin x\,dx=\int_0^{\pi/2}\log\cos x\,dx$,

$$2J = \int_0^{\pi/2}\log\left(\sin x\cos x\right)dx = \int_0^{\pi/2}\log\frac{\sin 2x}{2}\,dx = \int_0^{\pi/2}\log\sin 2x\,dx – \frac{\pi}{2}\log 2$$

Substituting $u=2x$ in that first integral gives $\frac12\int_0^{\pi}\log\sin u\,du$, which is $\frac12\cdot 2J = J$. So $2J = J – \frac\pi2\log 2$, giving $J = -\frac{\pi}{2}\log 2$. Therefore $2I = \pi\cdot 2J = -\pi^2\log 2$.

Two separate self-referential steps. The reflection removes the $x$; the doubling identity then makes $J$ appear on both sides of its own equation. Neither is a substitution in the ordinary sense, and this question is the clearest example in the unit of an integral solved by identity rather than by antiderivative. The book prints the answer as $\frac{\pi^2}{2}\log\frac12$, which is the same number.

Question 47

$$\int_{-\pi/4}^{\pi/4}\log\left(\sin x+\cos x\right)dx$$

Solution. The symmetric limits suggest testing parity, but $\log(\sin x+\cos x)$ is neither even nor odd. Instead, combine the two terms into one:

$$\sin x+\cos x = \sqrt2\,\cos\left(x-\frac\pi4\right)$$

Substituting $u=x-\frac\pi4$ shifts the interval to $\left[-\frac\pi2,\,0\right]$ and splits the logarithm:

$$\int_{-\pi/2}^{0}\left[\log\sqrt2 + \log\cos u\right]du = \frac{\pi}{2}\cdot\frac{\log 2}{2} + \int_{-\pi/2}^{0}\log\cos u\,du$$

$\log\cos u$ is even, so the second integral equals $\int_0^{\pi/2}\log\cos u\,du$, which is the $J$ computed in Question 46 and equals $-\frac\pi2\log 2$. Hence the value is $\frac\pi4\log 2 – \frac\pi2\log 2$.

This question depends on the previous one. $\int_0^{\pi/2}\log\cos u\,du$ is not a standard form to be quoted from memory — it is derived in Question 46, and the exemplar has placed the two adjacently for that reason. Solving 47 without having done 46 means deriving the same result twice.

Common mistakes

• Splitting $x^4-x^2-12$ in Questions 35 into linear factors. It factorises as $(x^2-4)(x^2+3)$, and the second factor has no real roots, so a four-term decomposition in $x$ cannot be completed. Treat the denominator as a quadratic in $x^2$ and stop there.
• Writing the Question 36 answer without noticing it fails at $a=b$. The $\frac{1}{a^2-b^2}$ out front is not a blemish; it is telling you the method assumed $a^2\neq b^2$ when it divided.
• Splitting the fraction in Question 39. $\frac{1+x+x^2}{1+x^2}=1+\frac{x}{1+x^2}$ is a true identity and a dead end, because the split destroys the $f+f’$ pattern that makes the integral elementary at all.
• Dropping the absolute values in Question 41 without checking the interval. Here $\sqrt{\cos^2\frac x2}=\cos\frac x2$ is legitimate because $\frac x2\in\left[\frac\pi6,\frac\pi4\right]$. On a different interval the same step would flip a sign.
• Attacking Question 42 by parts before linearising. Integration by parts on $e^{-3x}\cos^3x$ generates terms in $\cos^2x\sin x$ and never returns to the original integral, so the loop never closes. Turn $\cos^3x$ into multiple angles first and every piece becomes standard.
• Using partial fractions on $\frac{x^2}{1+2x}$ in Question 45. The numerator’s degree is not lower than the denominator’s, so long division comes first. This is the most common single error in the band.
• Forgetting to change the limits in Questions 41, 44 and 47. All three substitute inside a definite integral, and Question 47’s shift to $\left[-\frac\pi2,0\right]$ is the one most often missed, because a shift feels less like a substitution than a $\tan$ or a $\sin$ does.
• Quoting $\int_0^{\pi/2}\log\sin x\,dx=-\frac\pi2\log 2$ in Question 46 as a known result. In an examination it must be derived, and the derivation — doubling the angle until $J$ reappears on both sides — is most of the marks.

Practise next

• Exemplar Objective and Fill in the Blanks Q48–63, the last band of the unit, where several of the methods above are tested in a single line each.
• Exemplar Short Answer Q18–34, if the definite-integral properties used in Questions 37, 46 and 47 were the difficulty here — that page introduces them on gentler examples.
• NCERT textbook Exercise 7.5 for partial fractions, which Questions 35, 36, 38 and 45 all rest on.
• NCERT textbook Exercise 7.10 for the properties of definite integrals — it is the textbook exercise that Questions 37, 46 and 47 extend.
• NCERT textbook Miscellaneous Exercise on Integrals, the hardest set in the textbook itself, and the closest thing in the book to the difficulty of this page.