NCERT Class 12 Mathematics — Integrals, Exercise 7.7. All 11 questions solved.

Exercise 7.7 is the shortest in the chapter and the most formulaic. Every question is the square root of a quadratic, and there are exactly three standard forms — each derived by integration by parts with $1$ as the second function:

$$\int \sqrt{x^2 – a^2}\,dx = \frac{x}{2}\sqrt{x^2-a^2} – \frac{a^2}{2}\log\left|x + \sqrt{x^2-a^2}\right| + \mathrm{C}$$

$$\int \sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2+a^2}\right| + \mathrm{C}$$

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

Key insight. The sign inside the square root decides everything. A plus in front of $x^2$ gives a logarithm; a minus gives an arcsine. And the sign in front of the second term follows the sign inside the root: $\sqrt{x^2-a^2}$ subtracts, the other two add.

The only work is completing the square to identify $a$, and then substituting $t$ for the shifted variable. Get that right and the rest is bookkeeping.

Questions 1 to 9 — integrate the functions.

Question 1

$\sqrt{4 – x^2}$

Solution. Already in the third form with $a = 2$:

Question 2

$\sqrt{1 – 4x^2}$

Solution. Put $t = 2x$, so $dt = 2\,dx$ and the root becomes $\sqrt{1 – t^2}$ with $a = 1$:

$$\frac{1}{2}\left[\frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin^{-1}t\right]$$

Question 3

$\sqrt{x^2 + 4x + 6}$

Solution. Complete the square:

$$x^2 + 4x + 6 = (x+2)^2 + 2$$

so $t = x + 2$ and $a^2 = 2$. The second coefficient is $\tfrac{a^2}{2} = 1$:

Question 4

$\sqrt{x^2 + 4x + 1}$

Solution.

$$x^2 + 4x + 1 = (x+2)^2 – 3$$

This is the first form — note the minus inside — so $a^2 = 3$ and the logarithm term is subtracted:

Question 5

$\sqrt{1 – 4x – x^2}$

Solution. The $x^2$ is negative, so pull the minus out first:

$$1 – 4x – x^2 = 1 – \left(x^2 + 4x\right) = 5 – (x+2)^2$$

Third form, $a = \sqrt{5}$, so the second coefficient is $\tfrac52$:

Question 6

$\sqrt{x^2 + 4x – 5}$

Solution.

$$x^2 + 4x – 5 = (x+2)^2 – 9$$

First form, $a = 3$:

Question 7

$\sqrt{1 + 3x – x^2}$

Solution. Half of 3 is $\tfrac32$, so:

$$1 + 3x – x^2 = 1 + \frac{9}{4} – \left(x – \frac{3}{2}\right)^2 = \frac{13}{4} – \left(x – \frac{3}{2}\right)^2$$

Third form with $a = \tfrac{\sqrt{13}}{2}$, so $\tfrac{a^2}{2} = \tfrac{13}{8}$:

Question 8

$\sqrt{x^2 + 3x}$

Solution. There is no constant term, but the method is unchanged:

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

First form with $a = \tfrac32$, so $\tfrac{a^2}{2} = \tfrac98$:

Question 9

$\sqrt{1 + \dfrac{x^2}{9}}$

Solution. Bring the whole expression over a common denominator inside the root and take the constant out:

$$\sqrt{1 + \frac{x^2}{9}} = \frac{\sqrt{9 + x^2}}{3}$$

Now the second form with $a = 3$, all divided by 3:

$$\frac{1}{3}\left[\frac{x}{2}\sqrt{x^2+9} + \frac{9}{2}\log\left|x + \sqrt{x^2+9}\right|\right]$$

Questions 10 and 11 — choose the correct answer.

Question 10

$\displaystyle\int \sqrt{1 + x^2}\,dx$ is equal to

Solution. The second standard form with $a = 1$, so $\tfrac{a^2}{2} = \tfrac12$ in both places.

Option (B) is the trap: it treats $\sqrt{1+x^2}$ as though the power rule applied, which would need an $x$ outside from the chain rule. Differentiating (B) gives $2x\sqrt{1+x^2}$, not $\sqrt{1+x^2}$.

Question 11

$\displaystyle\int \sqrt{x^2 – 8x + 7}\,dx$ is equal to

Solution. Complete the square:

$$x^2 – 8x + 7 = (x-4)^2 – 9$$

First form with $t = x – 4$ and $a = 3$, so the logarithm’s coefficient is $-\tfrac{a^2}{2} = -\tfrac92$.

All four options differ only in that coefficient and its sign. (A) has the sign wrong — the minus inside the root forces a minus outside. (C) uses $3\sqrt2$, which comes from taking $a^2 = 18$ somewhere. (B) also mis-shifts the variable.

Common mistakes

• Using the wrong standard form. $\sqrt{x^2 – a^2}$ and $\sqrt{a^2 – x^2}$ look similar and give completely different answers — a logarithm versus an arcsine. Check which term carries the minus before writing anything.
• Getting the sign of the second term wrong. Only $\sqrt{x^2-a^2}$ subtracts. Question 11’s option (A) is exactly this error.
• Writing $a$ where $a^2$ belongs. The coefficient is $\tfrac{a^2}{2}$, so $a = 3$ gives $\tfrac92$, not $\tfrac32$. This is the single most common slip in the exercise.
• Treating the root as a power to be integrated directly. $\int\sqrt{1+x^2}\,dx$ is not $\tfrac23(1+x^2)^{3/2}$ — that would need an $x$ in the integrand. Question 10 offers it as option (B).
• Forgetting the extra constant from a substitution. In question 2, $t = 2x$ brings a factor of $\tfrac12$ outside; in question 9 the $\tfrac13$ must be carried through both terms.

Practise next

• Exercise 7.8 — definite integrals, evaluated with the Fundamental Theorem of Calculus.
• Exercise 7.4 — the same completing-the-square skill without the outer square root.