CBSE Class 12 Mathematics Inverse Trigonometric Functions Assignment Set 05

The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table:

\( \sin^{-1}x \) should not be condused with \( (\sin x)^{-1} \). In fact \( (\sin x)^{-1} = \frac{1}{\sin x} \) And similarly for other trigonometric functions

• The value of an inverse trigonometric functions which lies in its principal value branch is called the principal value of that inverse trigonometric functions.

For suitable values of domain, we have

• \( y = \sin^{-1} x \Rightarrow x = \sin y \)
• \( x = \sin y \Rightarrow y = \sin^{-1} x \)
• \( \sin(\sin^{-1} x) = x \)
• \( \sin^{-1}(\sin x) = x \)

Key Notes

• \( \sin^{-1}\frac{1}{x} = \mathrm{cosec}^{-1}x \)
• \( \cos^{-1}(-x) = \pi - \cos^{-1}x \)
• \( \cos^{-1}\frac{1}{x} = \sec^{-1}x \)
• \( \cot^{-1}(-x) = \pi - \cot^{-1}x \)
• \( \tan^{-1}\frac{1}{x} = \cot^{-1}x \)
• \( \sec^{-1}(-x) = \pi - \sec^{-1}x \)
• \( \sin^{-1}(-x) = -\sin^{-1}x \)
• \( \tan^{-1}(-x) = -\tan^{-1}x \)
• \( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \)
• \( \mathrm{cosec}^{-1}(-x) = -\mathrm{cosec}^{-1}x \)
• \( \mathrm{cosec}^{-1}x + \sec^{-1}x = \frac{\pi}{2} \)
• \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \)
• \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy} \)
• \( 2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2} \)
• \( \tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy} \)
• \( 2\tan^{-1}x = \sin^{-1}\frac{2x}{1+x^2} = \cos^{-1}\frac{1-x^2}{1+x^2} \)

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 :

\(\begin{cases} \displaystyle a \left[ \left( x + \frac{b}{2a} \right)^2 - \frac{\sqrt{4ac - b^2}}{2a} \right]^2 & \text{when } b^2 < 4ac. \\ \displaystyle a \left[ \left( x + \frac{b}{2a} \right)^2 - \left(\frac{\sqrt{b^2 - 4ac}}{2a}\right)^2 \right] & \text{when } b^2 > 4ac. \end{cases}\)

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

\(\int \sqrt{a^2 + x^2} \, dx, \quad \int \sqrt{a^2 - x^2} \, dx, \quad \int \sqrt{x^2 - a^2} \, dx\), 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) \(\int \sqrt{x^2 + 4x + 8} \; dx = \int \sqrt{x^2 + 4x + 4 + 4} \; dx\)
\(= \int \sqrt{(x + 2)^2 + (2)^2} \; dx\)
\(= \frac{x + 2}{2} \sqrt{x^2 + 4x + 8} + \frac{4}{2} \log \left| (x + 2) + \sqrt{x^2 + 4x + 8} \right| + c\)
\(= \frac{1}{2} (x + 2) \sqrt{x^2 + 4x + 8} + 2 \log \left| (x + 2) + \sqrt{x^2 + 4x + 8} \right| + c\)
Answer: \(\frac{1}{2} (x + 2) \sqrt{x^2 + 4x + 8} + 2 \log \left| (x + 2) + \sqrt{x^2 + 4x + 8} \right| + c\)
In simple words: Complete the square inside the square root and apply the standard integral formula for \(\sqrt{x^2 + a^2}\).

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

Solution : (ii) \(\int \sqrt{(x - 5)(7 - x)} \; dx = \int \sqrt{12x - 35 - x^2} \; dx\)
\(= \int \sqrt{-35 - (x^2 - 12x)} \; dx = \int \sqrt{-35 - (x^2 - 12x + 36 - 36)} \; dx\)
\(= \int \sqrt{(36 - 35) - (x - 6)^2} \; dx = \int \sqrt{1^2 - (x - 6)^2} \; dx\)
\(= \frac{x - 6}{2} \sqrt{(x - 5)(7 - x)} + \frac{1}{2} \sin^{-1} \left( \frac{x - 6}{1} \right) + c\)
Answer: \(\frac{x - 6}{2} \sqrt{(x - 5)(7 - x)} + \frac{1}{2} \sin^{-1} (x - 6) + c\)
In simple words: Multiply the binomials, rearrange into a quadratic expression with a negative leading term, complete the square, and apply the standard formula for \(\sqrt{a^2 - x^2}\).

Exam Tip: Pay careful attention to signs when factoring out a negative sign from the quadratic terms.

Solution : (iii) \(\int \sqrt{14x - 20 - 2x^2} \; dx = \sqrt{2} \int \sqrt{-10 + 7x - x^2} \; dx\)
\(= \sqrt{2} \int \sqrt{-10 - (x^2 - 7x)} \; dx = \sqrt{2} \int \sqrt{-10 - \left( x^2 - 7x + \frac{49}{4} - \frac{49}{4} \right)} \; dx\)
\(= \sqrt{2} \int \sqrt{-10 + \frac{49}{4} - \left( x - \frac{7}{2} \right)^2} \; dx = \sqrt{2} \int \sqrt{\left(\frac{3}{2}\right)^2 - \left(x - \frac{7}{2}\right)^2} \; dx\)
\(= \sqrt{2} \left[ \frac{x - \frac{7}{2}}{2} \sqrt{\left(\frac{3}{2}\right)^2 - \left(x - \frac{7}{2}\right)^2} + \frac{9}{8} \sin^{-1} \left( \frac{x - \frac{7}{2}}{\frac{3}{2}} \right) \right] + c\)
\(= \sqrt{2} \left[ \frac{2x - 7}{4} \sqrt{7x - 10 - x^2} + \frac{9}{8} \sin^{-1} \left( \frac{2x - 7}{3} \right) \right] + c\)
Answer: \(\frac{\sqrt{2}(2x - 7)}{4} \sqrt{7x - 10 - x^2} + \frac{9\sqrt{2}}{8} \sin^{-1} \left( \frac{2x - 7}{3} \right) + c\)
In simple words: Factor out the coefficient of \(x^2\) first, then complete the square inside the radical and integrate using standard results.

Exam Tip: Do not forget to distribute the constant factor (like \(\sqrt{2}\)) across all terms inside the bracket at the final stage.

Solution : (iv) \(\int \sqrt{4ax - x^2} \; dx = \int \sqrt{-(x^2 - 4ax + 4a^2 - 4a^2)} \; dx\)
\(= \int \sqrt{4a^2 - (x^2 - 4ax + 4a^2)} \; dx = \int \sqrt{(2a)^2 - (x - 2a)^2} \; dx\)
\(= \frac{x - 2a}{2} \sqrt{4ax - x^2} + \frac{4a^2}{2} \sin^{-1} \left( \frac{x - 2a}{2a} \right) + c\)
\(= \frac{1}{2} (x - 2a) \sqrt{4ax - x^2} + 2a^2 \sin^{-1} \left( \frac{x - 2a}{2a} \right) + c\)
Answer: \(\frac{1}{2} (x - 2a) \sqrt{4ax - x^2} + 2a^2 \sin^{-1} \left( \frac{x - 2a}{2a} \right) + c\)
In simple words: Complete the square for the terms involving \(x\) and apply the standard formula for \(\sqrt{a^2 - x^2}\).

Exam Tip: Recognize standard parameters quickly to save time during substitution.

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 \(I = \int (x - 3) \sqrt{x^2 + 4x + 3} \; dx\)
\(x - 3 = A(2x + 4) + B\)
\(\Rightarrow A = \frac{1}{2}, B = -5\)
\(\therefore I = \int \left[ \frac{1}{2}(2x + 4) \sqrt{x^2 + 4x + 3} - 5 \sqrt{x^2 + 4x + 3} \right] dx\)
\(= \frac{1}{3} (x^2 + 4x + 3)^{3/2} - 5 \int \sqrt{(x + 2)^2 - (1)^2} \; dx\)
\(= \frac{1}{3} (x^2 + 4x + 3)^{3/2} - 5 \left[ \frac{x + 2}{2} \sqrt{x^2 + 4x + 3} - \frac{1}{2} \log \left| (x + 2) + \sqrt{x^2 + 4x + 3} \right| \right] + c\)
Answer: \(\frac{1}{3} (x^2 + 4x + 3)^{3/2} - \frac{5(x + 2)}{2} \sqrt{x^2 + 4x + 3} + \frac{5}{2} \log \left| (x + 2) + \sqrt{x^2 + 4x + 3} \right| + c\)
In simple words: Express the linear factor as a constant times the derivative of the quadratic expression plus another constant, then split into two standard integrals.

Exam Tip: Always double check the derivative calculation when equating coefficients for A and B.

Solution : (ii) \(I = \int (3x + 5) \sqrt{2x^2 + 3x + 7} \; dx\)
\(3x + 5 = A(4x + 3) + B \implies 4A = 3 \text{ and } 3A + B = 5\)
\(\Rightarrow A = \frac{3}{4} \text{ and } B = \frac{11}{4}\)
\(I = \int \left[ \frac{3}{4}(4x + 3) \sqrt{2x^2 + 3x + 7} + \frac{11}{4} \sqrt{2} \sqrt{x^2 + \frac{3}{2}x + \frac{7}{2}} \right] dx\)
\(= \frac{1}{2} (2x^2 + 3x + 7)^{3/2} + \frac{11\sqrt{2}}{4} \int \sqrt{\left(x + \frac{3}{4}\right)^2 + \frac{7}{2} - \frac{9}{16}} \; dx\)
\(= \frac{1}{2} (2x^2 + 3x + 7)^{3/2} + \frac{11\sqrt{2}}{4} \int \sqrt{\left(x + \frac{3}{4}\right)^2 + \left(\frac{\sqrt{47}}{4}\right)^2} \; dx\)
Answer: \(\frac{1}{2} (2x^2 + 3x + 7)^{3/2} + \frac{11\sqrt{2}}{4} \left[ \frac{4x + 3}{8} \sqrt{2x^2 + 3x + 7} + \frac{47}{32} \log \left| \left(x + \frac{3}{4}\right) + \sqrt{x^2 + \frac{3}{2}x + \frac{7}{2}} \right| \right] + c\)
In simple words: Use the linear-derivative method to split the integral into a direct power integration and a standard radical integration.

Exam Tip: Take extra care with fractions when completing the square under coefficients greater than 1.