NCERT Class 12 Mathematics — Integrals, Exercise 7.10. All 21 questions solved.

This is the exercise that pays for the whole chapter. Almost none of these integrals can be done by finding an antiderivative — several of the integrands have no elementary antiderivative at all — and yet most of them come out in three lines. The properties used are:

$$\mathbf{P_2}: \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$$

$$\mathbf{P_3}: \int_a^b f(x)\,dx = \int_a^b f(a + b – x)\,dx \qquad \mathbf{P_4}: \int_0^a f(x)\,dx = \int_0^a f(a – x)\,dx$$

$$\mathbf{P_6}: \int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx \text{ if } f(2a-x) = f(x), \text{ and } 0 \text{ if } f(2a-x) = -f(x)$$

$$\mathbf{P_7}: \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx \text{ if } f \text{ is even}, \text{ and } 0 \text{ if } f \text{ is odd}$$

Key insight. $\mathbf{P_4}$ produces a second expression for the same number $I$. What you do with it decides the question. If the new integrand is the complement of the old one, add the two and the sum collapses — that is questions 1, 2, 3, 4 and 17. If the new integrand is the negative of the old one, you get $I = -I$, so $I = 0$ — that is questions 15 and 21. If it is the old one with a constant subtracted, you get $I = c – I$ — that is questions 8 and 19.

Three outcomes, one property. Recognising which of the three you are in is the entire skill this exercise trains.

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Question 1

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

Solution. By $\mathbf{P_4}$ with $a = \tfrac{\pi}{2}$, and using $\cos\left(\tfrac{\pi}{2} – x\right) = \sin x$:

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

Adding the two expressions for $I$ makes the Pythagorean identity available:

$$2I = \int_0^{\pi/2}\left(\sin^2 x + \cos^2 x\right) dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$$

Question 2

$\displaystyle\int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx$

Solution. There is no antiderivative to find here — the property is the only route. $\mathbf{P_4}$ swaps $\sin$ and $\cos$ throughout:

$$I = \int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx$$

The two integrands share a denominator, and their numerators add to it:

$$2I = \int_0^{\pi/2}\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx = \frac{\pi}{2}$$

Question 3

$\displaystyle\int_{0}^{\pi/2} \frac{\sin^{3/2} x\,dx}{\sin^{3/2} x + \cos^{3/2} x}$

Solution. Identical structure to question 2 — the exponent is irrelevant. $\mathbf{P_4}$ gives the complementary integrand, and the two add to $1$:

$$2I = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$$

Question 4

$\displaystyle\int_{0}^{\pi/2} \frac{\cos^5 x\,dx}{\sin^5 x + \cos^5 x}$

Solution. Again the same. Any integral of the form $\displaystyle\int_0^{\pi/2}\frac{f(\sin x)}{f(\sin x) + f(\cos x)}\,dx$ equals $\tfrac{\pi}{4}$, and questions 2, 3 and 4 are three instances of it.

Question 5

$\displaystyle\int_{-5}^{5} |x + 2|\,dx$

Solution. A modulus is a different formula on each side of its zero, so $\mathbf{P_2}$ must be used to split at $x = -2$:

$$|x+2| = \begin{cases} -(x+2), & -5 \le x < -2 \\ x+2, & -2 \le x \le 5\end{cases}$$

$$\int_{-5}^{-2}-(x+2)\,dx = \left[-\frac{(x+2)^2}{2}\right]_{-5}^{-2} = \frac92, \qquad \int_{-2}^{5}(x+2)\,dx = \frac{49}{2}$$

Question 6

$\displaystyle\int_{2}^{8} |x – 5|\,dx$

Solution. Split at $x = 5$, which sits inside $[2, 8]$:

$$\int_2^5 (5-x)\,dx + \int_5^8 (x-5)\,dx = \frac92 + \frac92$$

Both pieces are triangles of base $3$ and height $3$, which is why they are equal.

Question 7

$\displaystyle\int_{0}^{1} x(1-x)^n\,dx$

Solution. Expanding $(1-x)^n$ would need the binomial theorem and an $n$-term sum. $\mathbf{P_4}$ with $a = 1$ instead moves the power onto $x$, where it is trivial:

$$I = \int_0^1 (1-x)\big(1-(1-x)\big)^n dx = \int_0^1 (1-x)x^n\,dx = \int_0^1\left(x^n – x^{n+1}\right) dx$$

$$= \frac{1}{n+1} – \frac{1}{n+2}$$

Question 8

$\displaystyle\int_{0}^{\pi/4} \log(1 + \tan x)\,dx$

Solution. Apply $\mathbf{P_3}$ with $a + b = \tfrac{\pi}{4}$. The compound angle formula gives

$$\tan\left(\frac{\pi}{4} – x\right) = \frac{1 – \tan x}{1 + \tan x}, \qquad 1 + \tan\left(\frac{\pi}{4} – x\right) = \frac{2}{1 + \tan x}$$

Taking logarithms turns the quotient into a difference:

$$I = \int_0^{\pi/4}\left[\log 2 – \log(1+\tan x)\right] dx = \frac{\pi}{4}\log 2 – I$$

This is the “constant minus $I$” case, so $2I = \tfrac{\pi}{4}\log 2$.

Question 9

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

Solution. $\mathbf{P_4}$ with $a = 2$ moves the root onto the simpler variable:

$$I = \int_0^2 (2-x)\sqrt{x}\,dx = \int_0^2\left(2x^{1/2} – x^{3/2}\right) dx = \frac43 \cdot 2\sqrt2 – \frac25 \cdot 4\sqrt2$$

$$= 8\sqrt2\left(\frac13 – \frac15\right)$$

Substitution would also work here; the property is shorter.

Question 10

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

Solution. Expand the second logarithm using $\sin 2x = 2\sin x\cos x$ before touching any property:

$$2\log\sin x – \left(\log 2 + \log\sin x + \log\cos x\right) = \log\sin x – \log\cos x – \log 2$$

By $\mathbf{P_4}$, $\int_0^{\pi/2}\log\sin x\,dx = \int_0^{\pi/2}\log\cos x\,dx$, so the first two terms cancel — without either of them ever being evaluated. What survives is the constant:

$$I = -\int_0^{\pi/2}\log 2\,dx = -\frac{\pi}{2}\log 2$$

Question 11

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

Solution. $\sin^2(-x) = \sin^2 x$, so the integrand is even and $\mathbf{P_7}$ halves the work:

$$I = 2\int_0^{\pi/2}\sin^2 x\,dx = 2 \cdot \frac{\pi}{4}$$

using question 1’s result, which holds for $\sin^2$ as well as $\cos^2$.

Question 12

$\displaystyle\int_{0}^{\pi} \frac{x\,dx}{1 + \sin x}$

Solution. $\mathbf{P_3}$ with $a + b = \pi$ leaves the denominator unchanged, because $\sin(\pi – x) = \sin x$, and replaces $x$ by $\pi – x$ on top:

$$I = \int_0^\pi \frac{(\pi – x)\,dx}{1 + \sin x} \quad\Longrightarrow\quad 2I = \pi\int_0^\pi\frac{dx}{1+\sin x}$$

The $x$ has been removed entirely. For the remaining integral, multiply above and below by $1 – \sin x$:

$$\frac{1 – \sin x}{\cos^2 x} = \sec^2 x – \sec x\tan x, \qquad \int = \Big[\tan x – \sec x\Big]$$

Written as $\dfrac{\sin x – 1}{\cos x} = \dfrac{-\cos x}{1 + \sin x}$ this is continuous across $x = \tfrac{\pi}{2}$, and evaluates to $1 – (-1) = 2$. Hence $2I = 2\pi$.

Question 13

$\displaystyle\int_{-\pi/2}^{\pi/2} \sin^7 x\,dx$

Solution. $\sin^7(-x) = -\sin^7 x$: an odd power of an odd function is odd. By $\mathbf{P_7}$ the integral over a symmetric interval is zero.

Question 14

$\displaystyle\int_{0}^{2\pi} \cos^5 x\,dx$

Solution. Note the interval is $[0, 2\pi]$, so $a = \pi$ in $\mathbf{P_6}$. Since $\cos(2\pi – x) = \cos x$, the first branch applies:

$$I = 2\int_0^\pi\cos^5 x\,dx$$

Now apply $\mathbf{P_6}$ again on $[0, \pi]$, where $a = \tfrac{\pi}{2}$. Here $\cos(\pi – x) = -\cos x$, so $\cos^5(\pi – x) = -\cos^5 x$ and the second branch applies, giving $0$.

Question 15

$\displaystyle\int_{0}^{\pi/2} \frac{\sin x – \cos x}{1 + \sin x\cos x}\,dx$

Solution. $\mathbf{P_4}$ swaps $\sin$ and $\cos$. The denominator is symmetric in the two and does not change; the numerator reverses sign:

$$I = \int_0^{\pi/2}\frac{\cos x – \sin x}{1 + \cos x\sin x}\,dx = -I$$

The only number equal to its own negative is zero.

Question 16

$\displaystyle\int_{0}^{\pi} \log(1 + \cos x)\,dx$

Solution. $\mathbf{P_4}$ with $a = \pi$ gives $\cos(\pi – x) = -\cos x$:

$$I = \int_0^\pi \log(1 – \cos x)\,dx$$

Adding the two turns the product of logarithms’ arguments into a difference of squares:

$$2I = \int_0^\pi \log\left(1 – \cos^2 x\right) dx = \int_0^\pi \log\sin^2 x\,dx = 2\int_0^\pi\log\sin x\,dx$$

By $\mathbf{P_6}$ (with $\sin(\pi – x) = \sin x$) this is $4\int_0^{\pi/2}\log\sin x\,dx$, and that standard integral equals $-\tfrac{\pi}{2}\log 2$. So $2I = -2\pi\log 2$.

Question 17

$\displaystyle\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a – x}}\,dx$

Solution. The same pattern as question 2, now with a general upper limit. $\mathbf{P_4}$ exchanges $\sqrt{x}$ and $\sqrt{a-x}$, and adding gives

$$2I = \int_0^a 1\,dx = a$$

Question 18

$\displaystyle\int_{0}^{4} |x – 1|\,dx$

Solution. Split at $x = 1$ using $\mathbf{P_2}$:

$$\int_0^1 (1 – x)\,dx + \int_1^4 (x – 1)\,dx = \frac12 + \frac92$$

Question 19

Show that $\displaystyle\int_{0}^{a} f(x)g(x)\,dx = 2\int_{0}^{a} f(x)\,dx$, if $f$ and $g$ are defined as $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$.

Solution. Let $I = \displaystyle\int_0^a f(x)g(x)\,dx$. Applying $\mathbf{P_4}$ to the whole product:

$$I = \int_0^a f(a-x)\,g(a-x)\,dx$$

Now use the two hypotheses. The first says $f(a-x) = f(x)$, and the second says $g(a-x) = 4 – g(x)$:

$$I = \int_0^a f(x)\big[4 – g(x)\big]\,dx = 4\int_0^a f(x)\,dx – \int_0^a f(x)g(x)\,dx$$

The last term is $I$ itself, so $2I = 4\displaystyle\int_0^a f(x)\,dx$, which is the required result. This is the “constant minus $I$” case again, exactly as in question 8 — the constant here is $4\int_0^a f$ rather than a number.

Choose the correct answer in Exercises 20 and 21.

Question 20

The value of $\displaystyle\int_{-\pi/2}^{\pi/2}\left(x^3 + x\cos x + \tan^5 x + 1\right) dx$ is

Solution. Test each term for parity over the symmetric interval:

• $x^3$ is odd.
• $x\cos x$ is odd $\times$ even, so odd.
• $\tan^5 x$ is an odd power of an odd function, so odd.
• $1$ is even.

By $\mathbf{P_7}$ the first three contribute nothing, and only the constant survives:

$$\int_{-\pi/2}^{\pi/2} 1\,dx = \pi$$

Option (A) is the trap for anyone who checks parity on the first three terms and stops there.

Question 21

The value of $\displaystyle\int_{0}^{\pi/2}\log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right) dx$ is

Solution. $\mathbf{P_4}$ interchanges $\sin x$ and $\cos x$, which inverts the fraction inside the logarithm:

$$I = \int_0^{\pi/2}\log\left(\frac{4 + 3\cos x}{4 + 3\sin x}\right) dx = -\int_0^{\pi/2}\log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right) dx = -I$$

using $\log\tfrac1u = -\log u$. Hence $I = 0$ — and note that no property of the numbers $4$ and $3$ was used, so the same argument works for any constants.

Common mistakes

• Reaching for an antiderivative. Questions 2, 3, 4, 8, 16 and 21 have integrands whose antiderivatives are not elementary. Time spent hunting for one is time lost.
• Applying $\mathbf{P_4}$ and then not knowing what to do with the result. The property gives a second expression for $I$. Add it to the first, or set it equal, depending on whether the two integrands are complementary, opposite, or differ by a constant.
• Using $\mathbf{P_4}$ when the lower limit is not zero. $\mathbf{P_4}$ requires $[0, a]$. For $[a, b]$ use $\mathbf{P_3}$ with $f(a+b-x)$, as in questions 8 and 12.
• Integrating a modulus without splitting. $\int_{-5}^{5}|x+2|\,dx$ is not $\left|\int_{-5}^{5}(x+2)\,dx\right|$. Find where the bracket changes sign and split there.
• Assuming any symmetric interval gives zero. Question 20 has three odd terms and one even one; only the odd terms vanish. Check every term separately.
• Applying $\mathbf{P_6}$ with the wrong $a$. In question 14 the interval is $[0, 2\pi]$, so $a = \pi$, not $2\pi$. Reading $a$ off the upper limit instead of halving it is the usual slip.

Practise next

• Miscellaneous Exercise on Chapter 7 — mixes every technique in the chapter and does not tell you which one to use.
• Exercise 8.1 — areas under curves, where these definite integrals acquire a geometric meaning.