Complete solutions for Class 12 Maths (NCERT) Exercise 7.7. Master the integration of functions involving the square root of a quadratic expression ($\mathbf{\sqrt{x^2 \pm a^2}}$ and $\mathbf{\sqrt{a^2 – x^2}}$). Covers the essential technique of completing the square to transform expressions before applying the three standard formulas.
This exercise requires the use of the three special integration formulas for $\mathbf{\sqrt{\text{quadratic}}}$ functions. Completing the square is necessary for most problems.
Table of Contents
The three key formulas are:
- $\int \sqrt{a^2 – x^2} dx = \frac{x}{2} \sqrt{a^2 – x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$
- $\int \sqrt{x^2 – a^2} dx = \frac{x}{2} \sqrt{x^2 – a^2} – \frac{a^2}{2} \log\left|x + \sqrt{x^2 – a^2}\right| + C$
- $\int \sqrt{x^2 + a^2} dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \log\left|x + \sqrt{x^2 + a^2}\right| + C$
Part 1: Applying the Formulas (1-9)
1. $\int \sqrt{4 – x^2} dx = \int \sqrt{2^2 – x^2} dx$
Using Formula 1 with $a = 2$:
$$= \frac{x}{2} \sqrt{4 – x^2} + \frac{4}{2} \sin^{-1}\left(\frac{x}{2}\right) + C$$
$$= \mathbf{\frac{x}{2} \sqrt{4 – x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) + C}$$
2. $\int \sqrt{1 – 4x^2} dx$
Factor out 4: $\int \sqrt{4(\frac{1}{4} – x^2)} dx = 2 \int \sqrt{(\frac{1}{2})^2 – x^2} dx$.
Using Formula 1 with $a = 1/2$:
$$2 \left[ \frac{x}{2} \sqrt{\frac{1}{4} – x^2} + \frac{1/4}{2} \sin^{-1}\left(\frac{x}{1/2}\right) \right] + C$$
$$= x \sqrt{\frac{1 – 4x^2}{4}} + \frac{1}{4} \sin^{-1}(2x) + C$$
$$= \mathbf{\frac{x}{2} \sqrt{1 – 4x^2} + \frac{1}{4} \sin^{-1}(2x) + C}$$
3. $\int \sqrt{x^2 + 4x + 6} dx$
Complete the square: $x^2 + 4x + 6 = (x^2 + 4x + 4) + 2 = (x + 2)^2 + (\sqrt{2})^2$.
Let $u = x + 2$. The integral is $\int \sqrt{u^2 + (\sqrt{2})^2} du$.
Using Formula 3 with $a = \sqrt{2}$:
$$= \frac{u}{2} \sqrt{u^2 + 2} + \frac{2}{2} \log\left|u + \sqrt{u^2 + 2}\right| + C$$
$$= \mathbf{\frac{x + 2}{2} \sqrt{x^2 + 4x + 6} + \log\left|x + 2 + \sqrt{x^2 + 4x + 6}\right| + C}$$
4. $\int \sqrt{x^2 + 4x + 1} dx$
Complete the square: $x^2 + 4x + 1 = (x^2 + 4x + 4) – 3 = (x + 2)^2 – (\sqrt{3})^2$.
Let $u = x + 2$. The integral is $\int \sqrt{u^2 – (\sqrt{3})^2} du$.
Using Formula 2 with $a = \sqrt{3}$:
$$= \frac{u}{2} \sqrt{u^2 – 3} – \frac{3}{2} \log\left|u + \sqrt{u^2 – 3}\right| + C$$
$$= \mathbf{\frac{x + 2}{2} \sqrt{x^2 + 4x + 1} – \frac{3}{2} \log\left|x + 2 + \sqrt{x^2 + 4x + 1}\right| + C}$$
5. $\int \sqrt{1 – 4x – x^2} dx$
Complete the square: $1 – (x^2 + 4x) = 1 – (x^2 + 4x + 4 – 4) = 5 – (x + 2)^2 = (\sqrt{5})^2 – (x + 2)^2$.
Let $u = x + 2$. The integral is $\int \sqrt{(\sqrt{5})^2 – u^2} du$.
Using Formula 1 with $a = \sqrt{5}$:
$$= \frac{u}{2} \sqrt{5 – u^2} + \frac{5}{2} \sin^{-1}\left(\frac{u}{\sqrt{5}}\right) + C$$
$$= \mathbf{\frac{x + 2}{2} \sqrt{1 – 4x – x^2} + \frac{5}{2} \sin^{-1}\left(\frac{x + 2}{\sqrt{5}}\right) + C}$$
6. $\int \sqrt{x^2 + 4x – 5} dx$
Complete the square: $x^2 + 4x – 5 = (x^2 + 4x + 4) – 9 = (x + 2)^2 – 3^2$.
Let $u = x + 2$. The integral is $\int \sqrt{u^2 – 3^2} du$.
Using Formula 2 with $a = 3$:
$$= \frac{u}{2} \sqrt{u^2 – 9} – \frac{9}{2} \log\left|u + \sqrt{u^2 – 9}\right| + C$$
$$= \mathbf{\frac{x + 2}{2} \sqrt{x^2 + 4x – 5} – \frac{9}{2} \log\left|x + 2 + \sqrt{x^2 + 4x – 5}\right| + C}$$
7. $\int \sqrt{1 + 3x – x^2} dx$
Complete the square: $1 – (x^2 – 3x) = 1 – (x^2 – 3x + 9/4 – 9/4) = 1 + 9/4 – (x – 3/2)^2 = 13/4 – (x – 3/2)^2 = \left(\frac{\sqrt{13}}{2}\right)^2 – \left(x – \frac{3}{2}\right)^2$.
Let $u = x – 3/2$. The integral is $\int \sqrt{a^2 – u^2} du$ with $a = \sqrt{13}/2$.
Using Formula 1:
$$= \frac{u}{2} \sqrt{a^2 – u^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{u}{a}\right) + C$$
$$= \frac{x – 3/2}{2} \sqrt{1 + 3x – x^2} + \frac{13/4}{2} \sin^{-1}\left(\frac{x – 3/2}{\sqrt{13}/2}\right) + C$$
$$= \mathbf{\frac{2x – 3}{4} \sqrt{1 + 3x – x^2} + \frac{13}{8} \sin^{-1}\left(\frac{2x – 3}{\sqrt{13}}\right) + C}$$
8. $\int \sqrt{x^2 + 3x} dx$
Complete the square: $x^2 + 3x = x^2 + 3x + 9/4 – 9/4 = \left(x + \frac{3}{2}\right)^2 – \left(\frac{3}{2}\right)^2$.
Let $u = x + 3/2$. The integral is $\int \sqrt{u^2 – a^2} du$ with $a = 3/2$.
Using Formula 2:
$$= \frac{u}{2} \sqrt{u^2 – a^2} – \frac{a^2}{2} \log\left|u + \sqrt{u^2 – a^2}\right| + C$$
$$= \frac{x + 3/2}{2} \sqrt{x^2 + 3x} – \frac{9/4}{2} \log\left|x + 3/2 + \sqrt{x^2 + 3x}\right| + C$$
$$= \mathbf{\frac{2x + 3}{4} \sqrt{x^2 + 3x} – \frac{9}{8} \log\left|x + \frac{3}{2} + \sqrt{x^2 + 3x}\right| + C}$$
9. $\int \sqrt{1 + \frac{x^2}{9}} dx = \int \sqrt{\frac{9 + x^2}{9}} dx = \frac{1}{3} \int \sqrt{x^2 + 3^2} dx$
Using Formula 3 with $a = 3$:
$$\frac{1}{3} \left[ \frac{x}{2} \sqrt{x^2 + 9} + \frac{9}{2} \log\left|x + \sqrt{x^2 + 9}\right| \right] + C$$
$$= \mathbf{\frac{x}{6} \sqrt{x^2 + 9} + \frac{3}{2} \log\left|x + \sqrt{x^2 + 9}\right| + C}$$
Part 2: Multiple Choice Questions
10. $\int \sqrt{1 + x^2} dx$
Using Formula 3 with $a = 1$:
$$= \frac{x}{2} \sqrt{x^2 + 1} + \frac{1^2}{2} \log\left|x + \sqrt{x^2 + 1}\right| + C$$
$$= \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \log|x + \sqrt{1 + x^2}| + C$$
This matches option (A).
The correct answer is (A) $\frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \log|x + \sqrt{1 + x^2}| + C$.
11. $\int \sqrt{x^2 – 8x + 7} dx$
Complete the square: $x^2 – 8x + 7 = (x^2 – 8x + 16) – 9 = (x – 4)^2 – 3^2$.
Let $u = x – 4$. The integral is $\int \sqrt{u^2 – 3^2} du$.
Using Formula 2 with $a = 3$:
$$= \frac{u}{2} \sqrt{u^2 – 9} – \frac{9}{2} \log\left|u + \sqrt{u^2 – 9}\right| + C$$
Substitute $u = x – 4$ and $u^2 – 9 = x^2 – 8x + 7$:
$$= \frac{x – 4}{2} \sqrt{x^2 – 8x + 7} – \frac{9}{2} \log\left|x – 4 + \sqrt{x^2 – 8x + 7}\right| + C$$
This matches option (A).
The correct answer is (A) $\frac{1}{2} (x – 4) \sqrt{x^2 – 8x + 7} – \frac{9}{2} \log|x – 4 + \sqrt{x^2 – 8x + 7}| + C$.