NCERT Class 12 Mathematics — Integrals, Exercise 7.8. All 22 questions solved.

This is the first exercise in the chapter where the integral has limits, and it is deliberately easy: every integrand here is one you already know how to antidifferentiate from Exercises 7.1 to 7.7. What is new is the last step. The Fundamental Theorem of Calculus says that if $F$ is any antiderivative of $f$ on $[a, b]$, then

$$\int_a^b f(x)\,dx = F(b) – F(a)$$

Key insight. The constant of integration disappears. If you use $F(x) + \mathrm{C}$ instead of $F(x)$, the $\mathrm{C}$ cancels in the subtraction — so there is no $+\mathrm{C}$ on any answer in this exercise. Writing one is the clearest sign that a student is applying a memorised recipe rather than the theorem.

The second thing to watch is that $F$ must be an antiderivative throughout $[a, b]$. Every question here is arranged so that it is, but the habit of checking the interval matters later.

Evaluate the definite integrals in Exercises 1 to 20.

Question 1

$\displaystyle\int_{-1}^{1} (x + 1)\,dx$

Solution. An antiderivative is $F(x) = \dfrac{x^2}{2} + x$. The theorem then asks only for two evaluations:

$$F(1) – F(-1) = \left(\frac12 + 1\right) – \left(\frac12 – 1\right) = \frac32 + \frac12$$

Question 2

$\displaystyle\int_{2}^{3} \frac{1}{x}\,dx$

Solution. $F(x) = \log|x|$, and both limits are positive so the modulus can be dropped:

$$\log 3 – \log 2 = \log\frac{3}{2}$$

Question 3

$\displaystyle\int_{1}^{2} (4x^3 – 5x^2 + 6x + 9)\,dx$

Solution. Antidifferentiate term by term:

$$F(x) = x^4 – \frac{5x^3}{3} + 3x^2 + 9x$$

$$F(2) = 16 – \frac{40}{3} + 12 + 18 = \frac{98}{3}, \qquad F(1) = 1 – \frac53 + 3 + 9 = \frac{34}{3}$$

Keeping both as thirds makes the subtraction clean.

Question 4

$\displaystyle\int_{0}^{\pi/4} \sin 2x\,dx$

Solution. $F(x) = -\dfrac{\cos 2x}{2}$, because differentiating brings down a factor of $2$ from the chain rule that the $\tfrac12$ must cancel.

$$-\frac{\cos\frac{\pi}{2}}{2} + \frac{\cos 0}{2} = 0 + \frac12$$

Question 5

$\displaystyle\int_{0}^{\pi/2} \cos 2x\,dx$

Solution. $F(x) = \dfrac{\sin 2x}{2}$, so the value is $\tfrac12\left(\sin\pi – \sin 0\right)$.

Both terms vanish. Geometrically the curve $\cos 2x$ completes a full period’s worth of positive and negative area over $[0, \tfrac{\pi}{2}]$, and the two cancel exactly.

Question 6

$\displaystyle\int_{4}^{5} e^{x}\,dx$

Solution. $F(x) = e^x$, so the value is $e^5 – e^4$. Factor out the smaller power, which is how NCERT prints it:

Question 7

$\displaystyle\int_{0}^{\pi/4} \tan x\,dx$

Solution. $\int \tan x\,dx = \log|\sec x|$. At the upper limit $\sec\tfrac{\pi}{4} = \sqrt2$, and at the lower limit $\sec 0 = 1$:

$$\log\sqrt2 – \log 1 = \frac12\log 2$$

Question 8

$\displaystyle\int_{\pi/6}^{\pi/4} \operatorname{cosec} x\,dx$

Solution. Use $\int \operatorname{cosec} x\,dx = \log\left|\tan\dfrac{x}{2}\right|$. The limits halve to $\tfrac{\pi}{8}$ and $\tfrac{\pi}{12}$, and both of those tangents have exact surd values:

$$\tan\frac{\pi}{8} = \sqrt2 – 1, \qquad \tan\frac{\pi}{12} = 2 – \sqrt3$$

Subtracting two logarithms turns into a single logarithm of the ratio:

Question 9

$\displaystyle\int_{0}^{1} \frac{dx}{\sqrt{1 – x^2}}$

Solution. $F(x) = \sin^{-1} x$, so the value is $\sin^{-1}1 – \sin^{-1}0$.

The integrand is unbounded at $x = 1$, yet the integral is finite — the area under the curve converges even though the curve does not.

Question 10

$\displaystyle\int_{0}^{1} \frac{dx}{1 + x^2}$

Solution. $F(x) = \tan^{-1} x$, giving $\tan^{-1}1 – \tan^{-1}0$.

Question 11

$\displaystyle\int_{2}^{3} \frac{dx}{x^2 – 1}$

Solution. This is the standard form $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|$ with $a = 1$:

$$F(x) = \frac12\log\left|\frac{x-1}{x+1}\right|$$

$$F(3) – F(2) = \frac12\left(\log\frac{2}{4} – \log\frac{1}{3}\right) = \frac12\log\frac{3}{2}$$

Question 12

$\displaystyle\int_{0}^{\pi/2} \cos^2 x\,dx$

Solution. $\cos^2 x$ has no elementary antiderivative until you flatten the square with the double-angle identity $\cos^2 x = \dfrac{1 + \cos 2x}{2}$:

$$F(x) = \frac{x}{2} + \frac{\sin 2x}{4}$$

$$F\!\left(\frac{\pi}{2}\right) – F(0) = \frac{\pi}{4} + \frac{\sin\pi}{4} = \frac{\pi}{4}$$

Question 13

$\displaystyle\int_{2}^{3} \frac{x\,dx}{x^2 + 1}$

Solution. The numerator is half the derivative of the denominator, so $F(x) = \tfrac12\log(x^2+1)$:

$$\frac12\left(\log 10 – \log 5\right) = \frac12\log 2$$

Question 14

$\displaystyle\int_{0}^{1} \frac{2x + 3}{5x^2 + 1}\,dx$

Solution. Split the numerator into the part that matches the derivative of the denominator and the constant left over. Since $\dfrac{d}{dx}(5x^2+1) = 10x$, the $2x$ term integrates to a logarithm:

$$\int_0^1 \frac{2x}{5x^2+1}\,dx = \frac15\Big[\log(5x^2+1)\Big]_0^1 = \frac15\log 6$$

The constant term is an arctangent. Writing $5x^2 + 1 = 5\left(x^2 + \tfrac15\right)$ gives $a = \tfrac{1}{\sqrt5}$:

$$\int_0^1 \frac{3}{5x^2+1}\,dx = \frac35 \cdot \sqrt5 \Big[\tan^{-1}\!\left(\sqrt5 x\right)\Big]_0^1 = \frac{3}{\sqrt5}\tan^{-1}\sqrt5$$

Question 15

$\displaystyle\int_{0}^{1} x\,e^{x^2}\,dx$

Solution. Put $t = x^2$, so $dt = 2x\,dx$. The limits must change with the variable: $x = 0$ gives $t = 0$, and $x = 1$ gives $t = 1$.

$$\frac12\int_0^1 e^t\,dt = \frac12(e – 1)$$

Changing the limits is what lets you stop without resubstituting.

Question 16

$\displaystyle\int_{1}^{2} \frac{5x^2}{x^2 + 4x + 3}\,dx$

Solution. The degree of the numerator equals that of the denominator, so divide first:

$$\frac{5x^2}{x^2+4x+3} = 5 – \frac{20x + 15}{x^2+4x+3}$$

Now factor $x^2+4x+3 = (x+1)(x+3)$ and split the remainder into partial fractions. Setting $\dfrac{20x+15}{(x+1)(x+3)} = \dfrac{A}{x+1} + \dfrac{B}{x+3}$ and substituting $x = -1$ and $x = -3$ gives $A = -\tfrac52$ and $B = \tfrac{45}{2}$. So the integrand is

$$5 + \frac{5}{2(x+1)} – \frac{45}{2(x+3)}$$

Integrating over $[1, 2]$:

$$5 + \frac52\log\frac{3}{2} – \frac{45}{2}\log\frac{5}{4}$$

Question 17

$\displaystyle\int_{0}^{\pi/4} (2\sec^2 x + x^3 + 2)\,dx$

Solution. Three independent antiderivatives:

$$F(x) = 2\tan x + \frac{x^4}{4} + 2x$$

At $x = \tfrac{\pi}{4}$, $\tan x = 1$ and $\dfrac{x^4}{4} = \dfrac{\pi^4}{4 \cdot 256} = \dfrac{\pi^4}{1024}$. At $x = 0$ everything vanishes.

Question 18

$\displaystyle\int_{0}^{\pi} \left(\sin^2\frac{x}{2} – \cos^2\frac{x}{2}\right) dx$

Solution. Do not expand — recognise the identity. Since $\cos^2\theta – \sin^2\theta = \cos 2\theta$ with $\theta = \tfrac{x}{2}$, the bracket is exactly $-\cos x$:

$$-\int_0^\pi \cos x\,dx = -\Big[\sin x\Big]_0^\pi = -(0 – 0)$$

Question 19

$\displaystyle\int_{0}^{2} \frac{6x + 3}{x^2 + 4}\,dx$

Solution. Same split as question 14. The $6x$ matches the derivative $2x$ of the denominator up to a factor of $3$:

$$\int_0^2 \frac{6x}{x^2+4}\,dx = 3\Big[\log(x^2+4)\Big]_0^2 = 3(\log 8 – \log 4) = 3\log 2$$

The constant term is an arctangent with $a = 2$:

$$\int_0^2 \frac{3}{x^2+4}\,dx = \frac32\Big[\tan^{-1}\frac{x}{2}\Big]_0^2 = \frac32 \cdot \frac{\pi}{4} = \frac{3\pi}{8}$$

Question 20

$\displaystyle\int_{0}^{1} \left(x e^{x} + \sin\frac{\pi x}{4}\right) dx$

Solution. The first term needs integration by parts with $x$ as the first function:

$$\int x e^x\,dx = x e^x – e^x, \qquad \Big[x e^x – e^x\Big]_0^1 = (e – e) – (0 – 1) = 1$$

The second is a direct sine integral:

$$\int_0^1 \sin\frac{\pi x}{4}\,dx = -\frac{4}{\pi}\Big[\cos\frac{\pi x}{4}\Big]_0^1 = -\frac{4}{\pi}\left(\frac{1}{\sqrt2} – 1\right) = \frac{4}{\pi} – \frac{2\sqrt2}{\pi}$$

using $\dfrac{4}{\sqrt2} = 2\sqrt2$.

Choose the correct answer in Exercises 21 and 22.

Question 21

$\displaystyle\int_{1}^{\sqrt3} \frac{dx}{1 + x^2}$ equals

Solution. $F(x) = \tan^{-1}x$, so the value is

$$\tan^{-1}\sqrt3 – \tan^{-1}1 = \frac{\pi}{3} – \frac{\pi}{4} = \frac{\pi}{12}$$

Option (A) is what you get by forgetting the lower limit entirely — a reminder that a definite integral is a difference, not a single evaluation.

Question 22

$\displaystyle\int_{0}^{2/3} \frac{dx}{4 + 9x^2}$ equals

Solution. Pull out the coefficient of $x^2$ so the standard form applies:

$$\frac{1}{4 + 9x^2} = \frac{1}{9\left(x^2 + \frac49\right)}, \qquad a = \frac23$$

$$F(x) = \frac19 \cdot \frac{3}{2}\tan^{-1}\frac{3x}{2} = \frac16\tan^{-1}\frac{3x}{2}$$

At $x = \tfrac23$ the argument is exactly $1$, so the value is $\tfrac16 \cdot \tfrac{\pi}{4}$.

Option (D) is the answer if you drop the $\tfrac19$; option (B) if you drop the $\tfrac12$ from $a = \tfrac23$.

Common mistakes

• Writing $+\mathrm{C}$ on a definite integral. The constant cancels in $F(b) – F(a)$. An answer to this exercise that ends in $\mathrm{C}$ is wrong by definition, not just untidy.
• Substituting without changing the limits. In question 15, once you write $t = x^2$ the limits become $t = 0$ and $t = 1$. They happen to be the same numbers here, which is a trap — in Exercise 7.9 they will not be. Either change the limits or resubstitute, never neither.
• Forgetting the chain-rule factor. $\int \sin 2x\,dx$ is $-\tfrac12\cos 2x$, not $-\cos 2x$. Questions 4 and 5 both check this.
• Expanding what an identity would collapse. Question 18 becomes a two-line problem once you see $\sin^2\tfrac{x}{2} – \cos^2\tfrac{x}{2} = -\cos x$, and a long one if you expand each square separately.
• Skipping the division in question 16. Partial fractions only work on a proper fraction. With the numerator and denominator both quadratic you must divide first, or the algebra will not close.
• Losing the constant when completing a standard form. In question 22 the $\tfrac19$ outside and the $\tfrac{3}{2}$ from $\tfrac1a$ multiply to $\tfrac16$. Both of the wrong options are made by dropping one of them.

Practise next

• Exercise 7.9 — the same definite integrals, but where a substitution is needed and the limits genuinely change.
• Exercise 7.10 — the properties of definite integrals, which turn several of these into one-line arguments.