CBSE Class 12 Mathematics Integrals Assignment Set 04

CHAPTER - 7

INTEGRALS

I. Integrals of Type : \(\int \sqrt{ax^2 + bx + c} \, dx\)

In order to evaluate the above type of integral, we put \(ax^2 + bx + c\) in the form :

\(\left[ a \left( x + \frac{b}{2a} \right)^2 + \frac{\sqrt{4ac - b^2}}{2a} \right]\) when \(b^2 < 4ac\).

\(\left[ a \left( x + \frac{b}{2a} \right)^2 - \frac{\sqrt{b^2 - 4ac}}{2a} \right]\) when \(b^2 > 4ac\).

and the integral reduces to one of the following three forms:

\(\int \sqrt{a^2 + x^2} \, dx\), \(\int \sqrt{a^2 - x^2} \, dx\), \(\int \sqrt{x^2 - a^2} \, dx\), which can be evaluated by using standard formulae.

Example 1: Evaluate

(i) \(\int \sqrt{x^2 + 4x + 8} \, dx\)
(ii) \(\int \sqrt{(x - 5)(7 - x)} \, dx\)
(iii) \(\int \sqrt{14x - 20 - 2x^2} \, dx\)
(iv) \(\int \sqrt{4a - x^2} \, dx\)

Solution : (i) \(\int \sqrt{x^2 + 4x + 8} \, dx = \int \sqrt{x^2 + 4x + 4 + 4} \, dx\)
\(= \int \sqrt{(x + 2)^2 + (2)^2} \, dx\)
\(= \frac{x + 2}{2} \sqrt{x^2 + 4x + 8} + \frac{4}{2} \log \left| (x + 2) + \sqrt{x^2 + 4x + 8} \right| + c\)
\(= \frac{1}{2} (x + 2) \sqrt{x^2 + 4x + 8} + 2 \log \left| (x + 2) + \sqrt{x^2 + 4x + 8} \right| + c\)

(ii) \(\int \sqrt{(x - 5)(7 - x)} \, dx = \int \sqrt{12x - 35 - x^2} \, dx\)
\(= \int \sqrt{-35 - (x^2 - 12x)} \, dx = \int \sqrt{-35 - (x^2 - 12x + 36 - 36)} \, dx\)
\(= \int \sqrt{(36 - 35) - (x - 6)^2} \, dx = \int \sqrt{1^2 - (x - 6)^2} \, dx\)
\(= \frac{x - 6}{2} \sqrt{(x - 5)(7 - x)} + \frac{1}{2} \sin^{-1} \left( \frac{x - 6}{1} \right) + c\)

(iii) \(\int \sqrt{14x - 20 - 2x^2} \, dx = \sqrt{2} \int \sqrt{-10 + 7x - x^2} \, dx\)
\(= \sqrt{2} \int \sqrt{-10 - (x^2 - 7x)} \, dx = \sqrt{2} \int \sqrt{-10 - \left( x^2 - 7x + \frac{49}{4} - \frac{49}{4} \right)} \, dx\)
\(= \sqrt{2} \int \sqrt{-10 + \frac{49}{4} - \left( x - \frac{7}{2} \right)^2} \, dx = \sqrt{2} \int \sqrt{\left(\frac{3}{2}\right)^2 - \left(x - \frac{7}{2}\right)^2} \, dx\)
\(= \sqrt{2} \left[ \frac{x - \frac{7}{2}}{2} \sqrt{\left(\frac{3}{2}\right)^2 - \left(x - \frac{7}{2}\right)^2} + \frac{9}{8} \sin^{-1} \left( \frac{x - \frac{7}{2}}{\frac{3}{2}} \right) \right] + c\)
\(= \sqrt{2} \left[ \frac{2x - 7}{4} \sqrt{7x - 10 - x^2} + \frac{9}{8} \sin^{-1} \left( \frac{2x - 7}{3} \right) \right] + c\)

(iv) \(\int \sqrt{4ax - x^2} \, dx = \int \sqrt{-(x^2 - 4ax + 4a^2 - 4a^2)} \, dx\)
\(= \int \sqrt{4a^2 - (x^2 - 4ax + 4a^2)} \, dx = \int \sqrt{(2a)^2 - (x - 2a)^2} \, dx\)
\(= \frac{x - 2a}{2} \sqrt{4ax - x^2} + \frac{4a^2}{2} \sin^{-1} \left( \frac{x - 2a}{2a} \right) + c\)
\(= \frac{1}{2} (x - 2a) \sqrt{4ax - x^2} + 2a^2 \sin^{-1} \left( \frac{x - 2a}{2a} \right) + c\)

II. Integrals of Type : \(\int (px + q) \sqrt{ax^2 + bx + c} \, dx\)

Here, \(px + q\) is written as

\(px + q = A \left[ \frac{d}{dx} (ax^2 + bx + c) \right] + B\) and the values of \(A\) and \(B\) are determined by equating the coefficients of \(x\) and constant terms on both sides.

Then, writing

\(\int (px + q) \sqrt{ax^2 + bx + c} \, dx = A \int (2ax + b) \sqrt{ax^2 + bx + c} \, dx + B \int \sqrt{ax^2 + bx + c} \, dx\)
\(= \frac{2A}{3} (ax^2 + bx + c)^{3/2} + B \int \sqrt{ax^2 + bx + c} \, dx\)

The second part is evaluated as explained in above example.

Example 2: Evaluate

(i) \(\int (x - 3) \sqrt{x^2 + 4x + 3} \, dx\)
(ii) \(\int (3x + 5) \sqrt{2x^2 + 3x + 7} \, dx\)
(iii) \(\int (x - 4) \sqrt{4 + 3x - x^2} \, dx\)
(iv) \(\int (5x - 1) \sqrt{6 + 5x - 2x^2} \, dx\)

Solution :

(i) Let \(I = \int (x - 3) \sqrt{x^2 + 4x + 3} \, dx\)
\(x - 3 = A(2x + 4) + B\)
\(\Rightarrow A = \frac{1}{2}, B = -5\)
\(\therefore I = \frac{1}{2} \int (2x + 4) \sqrt{x^2 + 4x + 3} \, dx - 5 \int \sqrt{x^2 + 4x + 3} \, dx\)
\(= \frac{1}{3} (x^2 + 4x + 3)^{3/2} - 5 \int \sqrt{(x + 2)^2 - (1)^2} \, dx\)
\(= \frac{1}{3} (x^2 + 4x + 3)^{3/2} - 5 \left[ \frac{x + 2}{2} \sqrt{x^2 + 4x + 3} - \frac{1}{2} \log \left| (x + 2) + \sqrt{x^2 + 4x + 3} \right| \right] + c\)

(ii) \(I = \int (3x + 5) \sqrt{2x^2 + 3x + 7} \, dx\)
\(3x + 5 = A(4x + 3) + B \Rightarrow 4A = 3\) and \(3A + B = 5\)
\(\Rightarrow A = \frac{3}{4}\) and \(B = \frac{11}{4}\)
\(I = \int \left( \frac{3}{4}(4x + 3) \right) \sqrt{2x^2 + 3x + 7} \, dx + \frac{11}{4} \int \sqrt{2} \sqrt{x^2 + \frac{3}{2}x + \frac{7}{2}} \, dx\)
\(= \frac{1}{2} (2x^2 + 3x + 7)^{3/2} + \frac{11\sqrt{2}}{4} \int \sqrt{\left(x + \frac{3}{4}\right)^2 + \frac{7}{2} - \frac{9}{16}} \, dx\)
\(= \frac{1}{2} (2x^2 + 3x + 7)^{3/2} + \frac{11\sqrt{2}}{4} \int \sqrt{\left(x + \frac{3}{4}\right)^2 + \left(\frac{\sqrt{47}}{4}\right)^2} \, dx\)
\(= \frac{1}{2} (2x^2 + 3x + 7)^{3/2} + \frac{11\sqrt{2}}{4} \left[ \frac{x + \frac{3}{4}}{2} \sqrt{x^2 + \frac{3}{2}x + \frac{7}{2}} + \frac{47}{32} \log \left| \left(x + \frac{3}{4}\right) + \sqrt{x^2 + \frac{3}{2}x + \frac{7}{2}} \right| \right] + C\)
\(= \frac{1}{2} (2x^2 + 3x + 7)^{3/2} + \frac{11\sqrt{2}}{4} \left[ \frac{4x + 3}{8} \sqrt{x^2 + \frac{3}{2}x + \frac{7}{2}} + \frac{47}{32} \log \left| \frac{4x + 3}{4} + \sqrt{x^2 + \frac{3}{2}x + \frac{7}{2}} \right| \right] + C\)

(iii) \(I = \int (x - 4) \sqrt{4 + 3x - x^2} \, dx\)
\(x - 4 = A(3 - 2x) + B \Rightarrow A = -\frac{1}{2}, B = -\frac{5}{2}\)
\(\therefore I = \int -\frac{1}{2}(3 - 2x) \sqrt{4 + 3x - x^2} \, dx + \left(-\frac{5}{2}\right) \int \sqrt{4 + 3x - x^2} \, dx\)
\(= -\frac{1}{3} (4 - 3x - x^2)^{3/2} - \frac{5}{2} \left[ \frac{2x - 3}{4} \sqrt{4 + 3x - x^2} + \frac{25}{8} \sin^{-1} \left( \frac{2x - 3}{5} \right) \right] + C\)

(iv) \(\therefore I = \int (5x - 1) \sqrt{6 + 5x - 2x^2} \, dx\)
\(5x - 1 = A(5 - 4x) \Rightarrow A = -\frac{5}{4}, B = \frac{21}{4}\)
\(\therefore I = \int \left(-\frac{5}{4}(5 - 4x)\right) \sqrt{6 + 5x - 2x^2} \, dx + \left(\frac{21}{4}\right) \int \sqrt{2} \sqrt{3 + \frac{5}{2}x - x^2} \, dx\)
\(= -\frac{5}{6} (6 + 5x - 2x^2)^{3/2} + \frac{21\sqrt{2}}{4} \int \sqrt{\left(\frac{73}{4}\right)^2 - \left(x - \frac{5}{4}\right)^2} \, dx\)
\(= -\frac{5}{6} (6 + 5x - 2x^2)^{3/2} + \frac{21\sqrt{2}}{4} \left[ \frac{x - \frac{5}{4}}{2} \sqrt{3 + \frac{5}{2}x - x^2} + \frac{73}{32} \sin^{-1} \left( \frac{4x - 5}{\sqrt{73}} \right) \right] + C\)
\(= -\frac{5}{6} (6 + 5x - 2x^2)^{3/2} + \frac{21\sqrt{2}}{4} \left[ \frac{4x - 5}{8} \sqrt{3 + \frac{5}{2}x - x^2} + \frac{73}{32} \sin^{-1} \left( \frac{4x - 5}{\sqrt{73}} \right) \right] + C\)

Note: The above integral can also be evaluated by substituting \(\tan \frac{x}{2} = t\).

Exercise 1.

Question 1. Evaluate:

(i) \(\int \sqrt{3x^2 + 4x + 1} \, dx\)
(ii) \(\int \sqrt{1 + 2x - 3x^2} \, dx\)
(iii) \(\int \sqrt{x^2 + 4x + 1} \, dx\)
(iv) \(\int \sqrt{3 - 2x - 2x^2} \, dx\)
(v) \(\int \sqrt{(x - 3)(5 - x)} \, dx\)
(vi) \(\int \sqrt{(2ax - x^2)} \, dx\)
Answer:
(i) \(\frac{1}{6} (3x + 2) \sqrt{3x^2 + 4x + 1} - \frac{\sqrt{3}}{18} \log \left| \left(x + \frac{2}{3}\right) + \sqrt{x^2 + \frac{4}{3}x + \frac{1}{3}} \right| + C\)
(ii) \(\frac{1}{6} (3x - 1) \sqrt{1 + 2x - 3x^2} + \frac{2\sqrt{3}}{9} \sin^{-1} \left( \frac{3x - 1}{2} \right) + C\)
(iii) \(\frac{1}{2} (x + 2) \sqrt{x^2 + 4x + 1} - \frac{3}{2} \log \left| (x + 2) + \sqrt{x^2 + 4x + 1} \right| + C\)
(iv) \(\sqrt{2} \left( \frac{2x + 1}{4} \right) \sqrt{\frac{3}{2} - x - x^2} + \frac{7}{8} \sin^{-1} \left( \frac{2x + 1}{\sqrt{7}} \right) + C\)
(v) \(\frac{1}{2} (x - 4) \sqrt{(x - 3)(5 - x)} + \frac{1}{2} \sin^{-1} (x - 4) + C\)
(vi) \(\frac{1}{2} (x - a) \sqrt{2ax - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x - a}{a} \right) + C\)
In simple words: These integrals of quadratic roots are solved by completing the square and applying standard integration formulae.

Exam Tip: Always make the coefficient of \(x^2\) equal to 1 inside the square root before completing the square.

Question 2. Evaluate:

(i) \(\int (2x + 3) \sqrt{x^2 + 4x + 3} \, dx\)
(ii) \(\int (2x - 5) \sqrt{2 + 3x - x^2} \, dx\)
(iii) \(\int (2x + 3) \sqrt{4x^2 + 5x + 6} \, dx\)
(iv) \(\int (2x - 5) \sqrt{x^2 + 4x + 3} \, dx\)
(v) \(\int (x + 1) \sqrt{1 - x - x^2} \, dx\)
(vi) \(\int (6x + 5) \sqrt{6 + 6x - 2x^2} \, dx\)
Answer:
(i) \(\frac{2}{3} (x^2 + 4x + 3)^{3/2} - \frac{x + 2}{2} \sqrt{x^2 - 4x + 3} + \frac{1}{2} \log \left| (x + 2) + \sqrt{x^2 + 4x + 3} \right| + C\)
(ii) \(-\frac{2}{3} (2 + 3x - x^2)^{3/2} - \left(\frac{2x - 3}{4}\right) \sqrt{2 + 3x - x^2} - \frac{17}{4} \sin^{-1} \left( \frac{2x - 3}{\sqrt{17}} \right) + C\)
(iii) \(\frac{1}{6} (4x^2 + 5x + 6)^{3/2} + \frac{7}{16} (8x + 5) \sqrt{x^2 + \frac{5}{4}x + \frac{3}{2}} + \frac{497}{256} \log \left| \frac{8x + 5}{8} + \sqrt{x^2 + \frac{5}{4}x + \frac{3}{2}} \right| + C\)
(iv) \(\frac{2}{3} (x^2 - 4x + 3)^{3/2} - \frac{1}{2}(x - 2)\sqrt{x^2 - 4x + 3} + \frac{1}{2}\log|(x - 2) + \sqrt{x^2 - 4x + 3}| + C\)
(v) \(-\frac{1}{3}(1 - x - x^2)^{3/2} + \frac{1}{4}\left(x + \frac{1}{2}\right)\sqrt{1 - x - x^2} + \frac{5}{16}\sin^{-1}\left(\frac{2x + 1}{\sqrt{5}}\right) + C\)
(vi) \(-(6 + x - 2x^2)^{3/2} + \frac{13}{16}(4x - 1)\sqrt{6 + x - 2x^2} + \frac{637}{32\sqrt{2}}\sin^{-1}\left(\frac{4x - 1}{7}\right) + C\)
In simple words: Express the linear factor as the derivative of the quadratic expression plus a constant to split into simpler integrals.

Exam Tip: Set \(linear = A(\text{derivative of quadratic}) + B\), find \(A\) and \(B\) by comparing coefficients, and integrate term by term.