ML Aggarwal Class 12 Maths Solutions Section A Chapter 09 Integration

9.1 INTEGRATION

When \( \frac{d}{dx} (F(x)) = f(x) \), then \( \int f(x) dx = F(x) + C \).

The function \( F(x) \) is referred to as the anti-derivative, primitive, or integral of the function \( f(x) \), and \( C \) is called the constant of integration or arbitrary constant.

The process of finding functions whose derivative is given is known as anti-differentiation or integration.

9.1.1 Some Elementary Standard Integrals

9.1.2 Four Standard Theorems

Theorem 1. \( \frac{d}{dx} \int f(x) dx = f(x) \)

Theorem 2. \( \int \alpha f(x) dx = \alpha \int f(x) dx \), for all \( \alpha \in \mathbb{R} \)

Theorem 3. \( \int (f_1(x) + f_2(x) - f_3(x) + ...) dx = \int f_1(x) dx + \int f_2(x) dx - \int f_3(x) dx + ... \)

Theorem 4. \( \int f'(g(x)) g'(x) dx = f(g(x)) + C \)

ILLUSTRATIVE EXAMPLES

Example 1. Evaluate the following integrals:
(i) \( \int \left( x - \frac{1}{x} \right)^3 dx \)
(ii) \( \int \frac{(a^x + b^x)^2}{a^x b^x} dx \)
Answer:
(i) Expanding the expression, we have:
\( \int \left( x - \frac{1}{x} \right)^3 dx = \int \left( x^3 - 3x - \frac{1}{x^3} - 3x \left(x - \frac{1}{x}\right) \right) dx \)
\( = \int \left( x^3 - x^{-3} - 3x + \frac{3}{x} \right) dx \)
\( = \frac{x^4}{4} + \frac{x^{-2}}{-2} - \frac{3x^2}{2} + 3 \log |x| + C \)
\( = \frac{x^4}{4} + \frac{1}{2x^2} - \frac{3x^2}{2} + 3 \log |x| + C \)

(ii) Breaking down the numerator and simplifying:
\( \int \frac{(a^x + b^x)^2}{a^x b^x} dx = \int \frac{a^{2x} + b^{2x} + 2a^x b^x}{a^x b^x} dx = \int \left[ \left(\frac{a}{b}\right)^x + \left(\frac{b}{a}\right)^x + 2 \right] dx \)
\( = \frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}} + \frac{\left(\frac{b}{a}\right)^x}{\log \frac{b}{a}} + 2x + C, a \neq b \)

When \( a = b \):
\( \int \frac{(a^x + a^x)^2}{a^x \cdot a^x} dx = \int \frac{(2a^x)^2}{a^{2x}} dx = \int 4 dx = 4x + C \)
In simple words: Expand brackets, group like terms, then use the standard integral formulas for powers and logarithms. When the base values match, the expression simplifies to a constant times x.

Exam Tip: Always expand algebraic expressions fully before integrating - this makes applying standard integral rules straightforward and reduces errors from trying to integrate complex forms directly.

Example 2. Evaluate the following integrals:
(i) \( \int 9^{\log_3 x} dx \)
(ii) \( \int e^{3 \log x} (x - 4) dx \)
(iii) \( \int a^{\log x} dx \)
Answer:
(i) Converting the base using exponential-logarithm properties:
\( \int 9^{\log_3 x} dx = \int (3^2)^{\log_3 x} dx = \int 3^{2 \log_3 x} dx = \int 3^{\log_3 x^2} dx = \int x^2 dx = \frac{x^3}{3} + C = \frac{1}{3}x^3 + C \)

(ii) Simplifying the exponential term:
\( \int e^{3 \log x} (x - 4) dx = \int e^{\log x^3} (x - 4) dx = \int x^3 \cdot x^{-4} dx = \int \frac{1}{x} dx = \log x + C \)

(iii) First, observe that \( a^{\log x} = x^{\log a} \). Taking the logarithm of both sides confirms this identity. Therefore:
\( \int a^{\log x} dx = \int x^{\log a} dx = \frac{x^{\log a + 1}}{\log a + 1} + C \)
In simple words: Use the property that an exponential with a logarithm inside can be rewritten as a power. Convert to that form first, then apply standard power integration rules.

Exam Tip: Remember the key identity \( a^{\log_b x} = x^{\log_b a} \) - recognising when to apply this transforms difficult-looking expressions into manageable ones.

Example 3. Evaluate the following integrals:
(i) \( \int \tan^2 x dx \)
(ii) \( \int \sqrt{1 - \sin 2x} dx \)
Answer:
(i) Using the trigonometric identity \( \tan^2 x = \sec^2 x - 1 \):
\( \int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x + C \)

(ii) Rewriting the expression under the square root using the identity \( 1 - \sin 2x = \cos^2 x + \sin^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2 \):
\( \int \sqrt{1 - \sin 2x} dx = \int \sqrt{\cos^2 x + \sin^2 x - 2 \sin x \cos x} dx = \int (\cos x - \sin x) dx = \sin x - (-\cos x) + C = \sin x + \cos x + C \)
In simple words: Look for trigonometric identities that can simplify the expression. A perfect square under a square root sign can often be simplified directly, and standard trigonometric integrals then apply.

Exam Tip: Always try to rewrite complex trigonometric expressions using standard identities like \( \sin^2 x + \cos^2 x = 1 \) or \( 1 + \tan^2 x = \sec^2 x \) before integrating.

Example 4. Evaluate the following integrals:
(i) \( \int \frac{1}{1 + \sec x} dx \)
(ii) \( \int \frac{\sin x}{1 + \sin x} dx \)
Answer:
(i) Rationalising the denominator:
\( \int \frac{1}{1 + \sec x} dx = \int \frac{1}{1 + \sec x} \cdot \frac{1 - \sec x}{1 - \sec x} dx = \int \frac{1 - \sec x}{\sec^2 x - 1} dx = \int \frac{\sec x - 1}{\sec^2 x - 1} dx \)
\( = \int \frac{\sec x - 1}{\tan^2 x} dx = \int \left( \frac{\sec x}{\tan^2 x} - \frac{1}{\tan^2 x} \right) dx = \int (\csc x \cot x - \csc^2 x) dx \)
\( = \int \csc x \cot x dx - \int (\csc^2 x - 1) dx = -\csc x - (-\cot x - x) + C = -\csc x + \cot x + x + C \)

(ii) Splitting the fraction strategically:
\( \int \frac{\sin x}{1 + \sin x} dx = \int \frac{(1 + \sin x) - 1}{1 + \sin x} dx = \int 1 dx - \int \frac{1}{1 + \sin x} dx \)
\( = x - \int \frac{1}{1 + \sin x} \cdot \frac{1 - \sin x}{1 - \sin x} dx = x - \int \frac{1 - \sin x}{\cos^2 x} dx \)
\( = x - \int (\sec^2 x - \sec x \tan x) dx = x - (\tan x - \sec x) + C = x - \tan x + \sec x + C \)
In simple words: When a trigonometric denominator seems tricky, try rationalising or rewriting the numerator to split the fraction into simpler parts. This often reveals standard integral forms.

Exam Tip: For integrals with \( 1 + \sin x \) or \( 1 + \cos x \) in the denominator, multiplying by the conjugate (or strategically rearranging the numerator) usually breaks the problem into manageable standard forms.

Example 5. Evaluate the following integrals:
(i) \( \int \cos^{-1}(\sin x) dx \)
(ii) \( \int \tan^{-1}(\sec x + \tan x) dx \)
(iii) \( \int \tan^{-1} \left( \frac{1 - \sin x}{1 + \sin x} \right) dx \)
Answer:
(i) Using the identity \( \sin x = \cos \left( \frac{\pi}{2} - x \right) \):
\( \int \cos^{-1}(\sin x) dx = \int \cos^{-1} \left( \cos \left( \frac{\pi}{2} - x \right) \right) dx = \int \left( \frac{\pi}{2} - x \right) dx = \frac{\pi}{2} x - \frac{x^2}{2} + C \)

(ii) Rewriting the argument of the inverse tangent:
\( \int \tan^{-1}(\sec x + \tan x) dx = \int \tan^{-1} \left( \frac{1 + \sin x}{\cos x} \right) dx = \int \tan^{-1} \left( \frac{1 - \cos(\pi/2 - x) + \sin(\pi/2 - x)}{...} \right) dx \)
After simplification (using half-angle formulas):
\( = \int \tan^{-1} \left( \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right) dx = \int \left( \frac{\pi}{4} + \frac{x}{2} \right) dx = \frac{\pi}{4} x + \frac{x^2}{4} + C = \frac{1}{4}(\pi + x) + C \)

(iii) Using half-angle relationships:
\( \int \tan^{-1} \left( \frac{1 - \sin x}{1 + \sin x} \right) dx = \int \tan^{-1} \left( \frac{1 - \cos(\pi/2 - x)}{1 + \cos(\pi/2 - x)} \right) dx = \int \tan^{-1} \left( \tan \left( \frac{\pi}{4} - \frac{x}{2} \right) \right) dx \)
\( = \int \left( \frac{\pi}{4} - \frac{x}{2} \right) dx = \frac{\pi}{4} x - \frac{x^2}{4} + C \)
In simple words: When inverse trigonometric functions appear, look for angles or expressions inside that match standard angle identities. Once you recognise the pattern, the inverse function cancels and you are left with a simple variable to integrate.

Exam Tip: Inverse trigonometric integrals almost always simplify if you spot a hidden angle (like \( \frac{\pi}{4} + \frac{x}{2} \)) inside the inverse function - work backwards from standard angle formulas to find these patterns.

EXERCISE 9.1

Evaluate the following (1 to 16) integrals:

Question 1. (i) \( \int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right) dx \)
(ii) \( \int \frac{3 + 5t - 7t^2}{\sqrt{t}} dt \)
Answer:
(i) \( \frac{x^2}{2} + \log |x| - 2x + C \)
(ii) \( 6\sqrt{t} + \frac{10}{3}t^{3/2} - \frac{14}{5}t^{5/2} + C \)
In simple words: Break the expression into separate terms and integrate each one. Remember that fractional powers (like \( x^{1/2} \)) follow the same power rule as whole number powers.

Exam Tip: Always split complicated fractions into individual terms first - this avoids careless errors and makes the application of power rules automatic.

Question 2. (i) \( \int (2x - 1)(x^2 + 1) dx \)
(ii) \( \int x^2 \left( 2x - \frac{1}{x} \right) dx \)
Answer:
(i) \( \frac{x^4}{2} - \frac{x^3}{3} + x^2 - x + C \)
(ii) \( \frac{4}{5}x^5 - \frac{4}{3}x^3 + x + C \)
In simple words: Multiply out the brackets completely before integrating. Each term will then fit a standard integral rule.

Exam Tip: Resist the temptation to integrate term-by-term before expanding - one moment of algebra saves multiple integration errors.

Question 3. (i) \( \int (25)^{\log_5 x} dx \)
(ii) \( \int 2^{4 \log x} dx \)
Answer:
(i) \( \frac{1}{3}x^3 + C \)
(ii) \( \frac{2}{3}x^{3/2} + C \)
In simple words: Convert the logarithmic-exponential form using the property \( a^{\log_b c} = c^{\log_b a} \). This transforms the expression into a simple power that is easy to integrate.

Exam Tip: Always convert exponential-logarithmic expressions to pure powers first - this one step often makes the entire problem trivial.

Question 4. (i) \( \int e^{2 \log x} (x - 3) dx \)
(ii) \( \int (e^{x \log a} + e^{a \log x} - a^{\log a}) dx \)
Answer:
(i) \( \log x + C \)
(ii) \( \frac{a^x}{\log a} + \frac{x^{a+1}}{a+1} - a^x \cdot x + C \)
In simple words: Simplify exponential and logarithmic expressions using the property \( e^{\log u} = u \) and \( a^{\log x} = x^{\log a} \). Once simplified, use standard integral rules.

Exam Tip: These problems test your comfort with logarithm and exponential identities as much as integration - spend time mastering the simplification step.

Question 5. (i) \( \int \frac{x^{10} - 1}{x - 1} dx \)
(ii) \( \int \frac{1}{1 - \cos 2x} dx \)
Answer:
(i) \( \frac{x^{10}}{10} + \frac{x^9}{9} + \frac{x^8}{8} + ... + x + C \)
(ii) \( -\frac{1}{2}\cot x + C \)
In simple words: For (i), divide out the polynomial to get a sum of powers. For (ii), use the identity \( 1 - \cos 2x = 2 \sin^2 x \) to simplify, then integrate the resulting trigonometric form.

Exam Tip: Polynomial long division and trigonometric identities are essential tools - when you see a ratio or double-angle expression, reach for these methods first.

Question 6. (i) \( \int \cot^2 x dx \)
(ii) \( \int \left( \tan^2 x - 7x^2 + \frac{1}{x} \right) dx \)
Answer:
(i) \( -\cot x - x + C \)
(ii) \( \tan x - x - \frac{7}{3}x^3 + \log |x| + C \)
In simple words: Use the identity \( \cot^2 x = \csc^2 x - 1 \) to convert to a standard form. Split multi-term expressions into separate integrals and handle each one with its matching rule.

Exam Tip: Always convert \( \tan^2 x \) and \( \cot^2 x \) using their Pythagorean identities - this transforms them into directly integrable forms instantly.

Question 7. (i) \( \int \frac{1}{1 + \cos x} dx \)
(ii) \( \int \frac{1}{1 + \sin x} dx \)
Answer:
(i) \( -\cot x + \csc x + C \)
(ii) \( \tan x - \sec x + C \)
In simple words: Multiply both numerator and denominator by a conjugate expression or an alternate form. For \( 1 + \cos x \), use \( 1 - \cos x \); for \( 1 + \sin x \), manipulate the fraction to expose standard trigonometric integrals.

Exam Tip: These are classic "rationalisation" integrals - memorising the standard forms saves time and avoids messy algebra during the exam.

Question 8. (i) \( \int \frac{\sin^2 x}{1 + \cos x} dx \)
(ii) \( \int \sqrt{1 + 2 \sin x} dx \)
Answer:
(i) \( x - \sin x + C \)
(ii) \( \sin x - \cos x + C \)
In simple words: For (i), use the identity \( \sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x) \), then cancel common factors. For (ii), recognise that the expression under the square root is a perfect square and simplify before integrating.

Exam Tip: Always look for ways to factor or simplify before integrating - a brief algebraic step often transforms a hard problem into an easy one.

Question 9. (i) \( \int (2 \tan x - 3 \cot x)^2 dx \)
(ii) \( \int \left( x^{3/2} - \cos \frac{x}{2} \right) dx \)
Answer:
(i) \( 4 \tan x - 9 \cot x - 25x + C \)
(ii) \( \frac{2}{5}x^{5/2} - \frac{1}{2}x - \frac{1}{2} \sin x + C \)
In simple words: Expand the squared expression and use trigonometric identities to convert to integrable forms. Split multi-term expressions and integrate each separately.

Exam Tip: When a trigonometric expression is squared, expand it first - this reveals which identities will be most helpful.

Question 10. (i) \( \int \left( \sqrt{x} - \sin \frac{x}{2} \cos \frac{x}{2} + 5 \right) dx \)
(ii) \( \int \frac{\cos 2x}{\sin^2 x \cos^2 x} dx \)
Answer:
(i) \( \frac{2}{3}x^{3/2} + \frac{1}{2} \cos x + 5x + C \)
(ii) \( -\cot x - \tan x + C \)
In simple words: For (i), use the double angle identity \( 2 \sin \frac{x}{2} \cos \frac{x}{2} = \sin x \) to simplify. For (ii), split the fraction and recognise standard trigonometric integrals.

Exam Tip: Double-angle identities like \( \sin 2x = 2 \sin x \cos x \) often hide inside complex expressions - spotting them saves significant work.

Question 11. (i) \( \int \frac{\sec x}{\sec x + \tan x} dx \)
(ii) \( \int \frac{\cos x - \cos 2x}{1 - \cos x} dx \)
Answer:
(i) \( \tan x - \sec x + C \)
(ii) \( 2 \sin x + x + C \)
In simple words: For (i), multiply numerator and denominator by a conjugate to simplify. For (ii), use the identity for \( \cos 2x \) to rewrite and factor the expression.

Exam Tip: When a denominator involves trig functions, multiplying by conjugates or using identities often reveals cancellations that simplify the problem dramatically.

Question 12. (i) \( \int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx \)
(ii) \( \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx \)
Answer:
(i) \( x + C \)
(ii) \( 2(\sin x + x \cos \alpha) + C \)
Hint (ii): \( \cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1) = 2(\cos^2 x - \cos^2 \alpha) \)
In simple words: For (i), notice that \( 1 + \sin 2x = (\sin x + \cos x)^2 \), so the square root simplifies. For (ii), apply the hint to factor the numerator and simplify the fraction.

Exam Tip: When \( \sin 2x \) or double-angle terms appear, expand them fully - they often form perfect squares or factorable expressions.

Question 13. (i) \( \int \frac{1}{\sin^2 x \cos^2 x} dx \)
(ii) \( \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx \)
Answer:
(i) \( \tan x - \cot x + C \)
(ii) \( \tan x - \cot x - 3x + C \)
In simple words: For (i), divide the numerator (which is 1) by each part of the denominator separately to get standard trig integrals. For (ii), use the factorisation \( a^3 + b^3 = (a + b)^3 - 3ab(a + b) \) to simplify the numerator first.

Exam Tip: Recognise algebraic identities like \( a^3 + b^3 \) and \( a^6 + b^6 \) - they often factorise in ways that simplify trigonometric expressions dramatically.

Question 14. (i) \( \int \cot^{-1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx \)
(ii) \( \int \tan^{-1} \left( \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} \right) dx \)
Answer:
(i) \( \frac{1}{2}x^2 + C \)
(ii) \( \frac{1}{2}x^2 + C \)
In simple words: Simplify the argument inside each inverse function using double-angle identities. The simplified forms will simplify to recognisable angles or expressions that can be cancelled with the inverse function.

Exam Tip: Inverse trigonometric integrals almost always simplify if you apply double-angle identities inside the function - this is the key first step for every such problem.

Question 15. (i) \( \int \sin^{-1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx \)
(ii) \( \int \cos^{-1} \left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) dx \)
Answer:
(i) \( x^2 + C \)
(ii) \( x^2 + C \)
In simple words: Recognise that the expressions inside are double-angle formulas - \( \sin 2x = \frac{2 \tan x}{1 + \tan^2 x} \) and \( \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \). Once you substitute these, the inverse functions cancel and you are left with simple integrals.

Exam Tip: Double-angle formulas in terms of \( \tan x \) are essential - memorise \( \sin 2x = \frac{2 \tan x}{1 + \tan^2 x} \) and \( \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \) for these problems.

Question 16. (i) \( \int \tan^{-1}(\csc x - \cot x) dx \)
(ii) \( \int \tan^{-1}(\csc x + \cot x) dx \)
Answer:
(i) \( \frac{1}{4}x^2 + C \)
(ii) \( \frac{\pi}{2}x - \frac{1}{4}x^2 + C \)
In simple words: Simplify the argument inside the inverse tangent by combining the cosecant and cotangent terms. These combinations often produce angle expressions that cancel with the inverse function.

Exam Tip: The combinations \( \csc x - \cot x \) and \( \csc x + \cot x \) often appear inside inverse functions - working with their tangent-half-angle equivalents is the fastest approach.

Question 17. If \( f'(x) = a \sin x + b \cos x \), \( f'(0) = 4 \), \( f(0) = 3 \), and \( f \left( \frac{\pi}{2} \right) = 5 \), find \( f(x) \).
Answer: Integrating the derivative, \( f(x) = \int (a \sin x + b \cos x) dx = -a \cos x + b \sin x + C \). From the condition \( f(0) = 3 \): \( -a + C = 3 \), so \( C = 3 + a \). Therefore, \( f(x) = -a \cos x + b \sin x + 3 + a \). From \( f'(0) = 4 \): \( a \sin 0 + b \cos 0 = 4 \), so \( b = 4 \). From \( f(\pi/2) = 5 \): \( -a \cos(\pi/2) + 4 \sin(\pi/2) + 3 + a = 5 \), which gives \( 0 + 4 + 3 + a = 5 \), so \( a = -2 \). Therefore, \( f(x) = 2 \cos x + 4 \sin x + 1 \).
In simple words: Integrate the derivative to find the general form, then use the three boundary conditions to determine the constants \( a \), \( b \), and \( C \).

Exam Tip: Always arrange conditions systematically - integrate first, then substitute each boundary condition one by one to avoid confusion when solving for multiple unknowns.

9.2 INTEGRATION BY SUBSTITUTION

Some functions integrate directly using standard formulas, while others cannot be integrated directly but can be changed into standard forms through a proper substitution - that is, by introducing a new variable. The method of evaluating an integral by converting it to a standard form through a substitution is called integration by substitution.

Remark. The method of substitution is essentially the use of Theorem 4, namely \( \int f'(g(x)) g'(x) dx = f(g(x)) + C \).

9.2.1 Theorem. If \( \int f(x) dx = F(x) + C \), then \( \int f(ax + b) dx = \frac{F(ax + b)}{a} + C, a \neq 0 \).

Proof. Set \( ax + b = t \), then \( a dx = dt \Rightarrow dx = \frac{1}{a} dt \). Therefore:
\( \int f(ax + b) dx = \int f(t) \cdot \frac{1}{a} dt = \frac{1}{a} \int f(t) dt = \frac{1}{a} F(t) + C = \frac{F(ax + b)}{a} + C, a \neq 0 \).

Important Rule. When the integral of a function in x is known, and x is multiplied by a constant with another constant added to the product, then the integral takes the same form but is divided by the coefficient of x.

Remark - Long Division. In a rational function, if the degree of the numerator is equal to or greater than that of the denominator, divide the numerator by the denominator until the degree of the remainder is less than that of the denominator. Remember: \( \frac{\text{numerator}}{\text{denominator}} = \text{quotient} + \frac{\text{remainder}}{\text{denominator}} \).

ILLUSTRATIVE EXAMPLES

Example 1. Evaluate the following integrals:
(i) \( \int \frac{1}{\sqrt{3x - 2}} dx \)
(ii) \( \int 10^{2x+1} dx \)
(iii) \( \int \sqrt{1 + \sin x} dx \)
Answer:
(i) Using substitution \( u = 3x - 2 \), so \( du = 3 dx \):
\( \int \frac{1}{\sqrt{3x - 2}} dx = \int (3x - 2)^{-1/2} dx = \frac{(3x - 2)^{1/2}}{\frac{1}{2} \cdot 3} + C = \frac{2}{3}\sqrt{3x - 2} + C \)

(ii) Using the formula for exponential integrals:
\( \int 10^{2x+1} dx = \frac{10^{2x+1}}{\log 10 \cdot 2} + C = \frac{10^{2x+1}}{2 \log 10} + C \)

(iii) Rewriting using the identity \( 1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right)^2 \):
\( \int \sqrt{1 + \sin x} dx = \int \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right) dx = \sin \frac{x}{2} - \cos \frac{x}{2} + C = 2 \left( \sin \frac{x}{2} - \cos \frac{x}{2} \right) + C \)
In simple words: For (i) and (ii), apply the substitution rule for linear expressions inside integrals - divide by the coefficient of x. For (iii), recognise a perfect square under the square root and simplify before integrating.

Exam Tip: The substitution rule for \( f(ax + b) \) is one of the quickest techniques - when you see a linear expression inside an integral, immediately think of dividing by the coefficient.

Example 2. Find all the primitives of the following functions:
(i) \( 5^{-2x} e^{-2x} \)
(ii) \( \frac{1}{\sqrt{2x + 3} + \sqrt{2x - 3}} \)
(iii) \( \frac{x^2}{1 - x} \)
Answer:
(i) Rewriting as a single base:
\( \int 5^{-2x} e^{-2x} dx = \int (5e)^{-2x} dx = \frac{(5e)^{-2x}}{\log(5e) \cdot (-2)} + C = -\frac{(5e)^{-2x}}{2 \log(5e)} + C \)

(ii) Rationalising the denominator:
\( \int \frac{1}{\sqrt{2x + 3} + \sqrt{2x - 3}} dx = \int \frac{1}{\sqrt{2x + 3} + \sqrt{2x - 3}} \cdot \frac{\sqrt{2x + 3} - \sqrt{2x - 3}}{\sqrt{2x + 3} - \sqrt{2x - 3}} dx \)
\( = \int \frac{\sqrt{2x + 3} - \sqrt{2x - 3}}{(2x + 3) - (2x - 3)} dx = \frac{1}{6} \int ((2x + 3)^{1/2} - (2x - 3)^{1/2}) dx \)
\( = \frac{1}{6} \left( \frac{(2x + 3)^{3/2}}{\frac{3}{2}} - \frac{(2x - 3)^{3/2}}{\frac{3}{2}} \right) + C = \frac{1}{18}((2x + 3)^{3/2} - (2x - 3)^{3/2}) + C \)

(iii) Using long division on the rational function:
\( \frac{x^2}{1 - x} = \frac{-(1 - x)x^2}{-(1 - x)} = -(x^2 + x) - \frac{1}{1 - x} \)
Wait, let me redo this by dividing \( x^2 \) by \( (1 - x) \):
\( \int \frac{x^2}{1 - x} dx = \int \left( -x - 1 + \frac{1}{1 - x} \right) dx = -\frac{x^2}{2} - x + \log|1 - x| + C = -\left( \frac{x^2}{2} + x + \log|1 - x| \right) + C \)
In simple words: For (i), combine the bases into a single term before integrating. For (ii), multiply by a conjugate to eliminate the sum in the denominator, then integrate each resulting power. For (iii), perform long division to break the improper fraction into simpler parts.

Exam Tip: Rationalisation and long division are essential algebraic techniques - when you see a denominator that is a sum of roots or an improper rational function, reach for these tools before attempting integration.

Example 3. Find all the anti-derivatives of the following functions:
(i) \( \frac{2x}{(2x + 1)^2} \)
(ii) \( (3x + 1)\sqrt{2x - 1} \)
(iii) \( \frac{x + 2}{\sqrt{3x + 1}} \)
Answer:
(i) Decomposing the numerator strategically:
\( \int \frac{2x}{(2x + 1)^2} dx = \int \frac{(2x + 1) - 1}{(2x + 1)^2} dx = \int \frac{1}{2x + 1} dx - \int (2x + 1)^{-2} dx \)
\( = \frac{\log|2x + 1|}{2} - \frac{(2x + 1)^{-1}}{2 \cdot (-1)} + C = \frac{1}{2} \left( \log|2x + 1| + \frac{1}{2x + 1} \right) + C \)

(ii) Expanding the product before integrating:
\( \int (3x + 1)\sqrt{2x - 1} dx = \int \left( \frac{3}{2}(2x - 1) + \frac{5}{2} \right) \sqrt{2x - 1} dx \)
\( = \frac{3}{2} \int (2x - 1)^{3/2} dx + \frac{5}{2} \int (2x - 1)^{1/2} dx \)
\( = \frac{3}{2} \cdot \frac{(2x - 1)^{5/2}}{5 \cdot 2 \cdot \frac{1}{2}} + \frac{5}{2} \cdot \frac{(2x - 1)^{3/2}}{3 \cdot 2 \cdot \frac{1}{2}} + C = \frac{3}{10}(2x - 1)^{5/2} + \frac{5}{6}(2x - 1)^{3/2} + C \)

(iii) Rewriting the numerator in terms of the expression under the square root:
\( \int \frac{x + 2}{\sqrt{3x + 1}} dx = \int \frac{\frac{1}{3}(3x + 1) + \frac{5}{3}}{\sqrt{3x + 1}} dx = \int \left( \frac{1}{3}(3x + 1)^{1/2} + \frac{5}{3}(3x + 1)^{-1/2} \right) dx \)
\( = \frac{1}{3} \cdot \frac{(3x + 1)^{3/2}}{3 \cdot \frac{3}{2}} + \frac{5}{3} \cdot \frac{(3x + 1)^{1/2}}{1 \cdot 3 \cdot \frac{1}{2}} + C = \frac{2}{27}(3x + 1)^{3/2} + \frac{10}{9}\sqrt{3x + 1} + C \)
In simple words: When the numerator and denominator have a related form, rewrite the numerator in terms of the denominator or its derivative. This often splits the fraction into simpler standard forms that integrate easily.

Exam Tip: This technique - expressing the numerator in terms of the denominator - is powerful and appears frequently. Practise recognising when to use it by looking for expressions that almost (but not quite) match the denominator or its derivative.

Example 4. Evaluate the following integrals:
(i) \( \int \sin^3 x dx \)
(ii) \( \int \cos^4 x dx \)
(iii) \( \int \sin 4x \cos 7x dx \)
Answer:
(i) Using the identity \( \sin 3x = 3 \sin x - 4 \sin^3 x \), we have \( \sin^3 x = \frac{3 \sin x - \sin 3x}{4} \):
\( \int \sin^3 x dx = \frac{3}{4}(- \cos x) - \frac{1}{4} \left( -\frac{\cos 3x}{3} \right) + C = \frac{1}{12}(\cos 3x - 9 \cos x) + C \)

(ii) Using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \), so \( \cos^4 x = \left( \frac{1 + \cos 2x}{2} \right)^2 \):
\( \int \cos^4 x dx = \frac{1}{4} \int (1 + 2 \cos 2x + \cos^2 2x) dx = \frac{1}{4} \int \left( 1 + 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) dx \)
\( = \frac{1}{8} \int (3 + 4 \cos 2x + \cos 4x) dx = \frac{1}{8} \left( 3x + 2 \sin 2x + \frac{\sin 4x}{4} \right) + C = \frac{1}{32}(12x + 8 \sin 2x + \sin 4x) + C \)

(iii) Using the product-to-sum formula \( 2 \sin A \cos B = \sin(A + B) + \sin(A - B) \):
\( \int \sin 4x \cos 7x dx = \frac{1}{2} \int (\sin 11x - \sin 3x) dx = \frac{1}{2} \left( -\frac{\cos 11x}{11} + \frac{\cos 3x}{3} \right) + C = \frac{1}{66}(-3 \cos 11x + 11 \cos 3x) + C \)
In simple words: For powers of sine or cosine, use reduction identities (like \( \sin 3x = 3 \sin x - 4 \sin^3 x \)) or double-angle formulas repeatedly until you reach integrable forms. For products of different trig functions, use product-to-sum formulas to break them into single sine or cosine terms.

Exam Tip: Trigonometric identities are essential tools - memorising reduction formulas and product-to-sum identities will save you from lengthy algebra and careless errors in these problems.

Example 5. Evaluate the following integrals:
(i) \( \int (\sin^6 x + \cos^6 x) dx \)
(ii) \( \int \frac{1 + \cos 4x}{\cot x - \tan x} dx \)
Answer:
(i) Using the identity \( a^3 + b^3 = (a + b)^3 - 3ab(a + b) \), with \( a = \sin^2 x \) and \( b = \cos^2 x \):
\( \int (\sin^6 x + \cos^6 x) dx = \int ((\sin^2 x)^3 + (\cos^2 x)^3) dx = \int ((\sin^2 x + \cos^2 x)^3 - 3 \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)) dx \)
\( = \int (1 - 3 \sin^2 x \cos^2 x) dx = \int \left( 1 - \frac{3}{4}(2 \sin x \cos x)^2 \right) dx = \int \left( 1 - \frac{3}{4} \sin^2 2x \right) dx \)
\( = \int \left( 1 - \frac{3}{8}(1 - \cos 4x) \right) dx = \frac{1}{8} \int (5 + 3 \cos 4x) dx = \frac{1}{8} \left( 5x + \frac{3 \sin 4x}{4} \right) + C = \frac{1}{32}(20x + 3 \sin 4x) + C \)

(ii) Simplifying the denominator and numerator:
\( \int \frac{1 + \cos 4x}{\cot x - \tan x} dx = \int \frac{2 \cos^2 2x}{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}} dx = \int \frac{2 \cos^2 2x \cdot \sin x \cos x}{\cos^2 x - \sin^2 x} dx = \int \frac{2 \cos^2 2x \cdot \sin x \cos x}{\cos 2x} dx \)
\( = \int \cos 2x \sin 2x dx = \frac{1}{2} \int 2 \sin 2x \cos 2x dx = \frac{1}{2} \int \sin 4x dx = \frac{1}{2} \cdot \left( -\frac{\cos 4x}{4} \right) + C = -\frac{1}{8} \cos 4x + C \)
In simple words: For (i), use algebraic factorisations on the powers of trigonometric functions to reveal simpler terms. For (ii), simplify the complex denominator by converting to sines and cosines, then use double-angle formulas to reduce further.

Exam Tip: When a trigonometric expression looks very complicated, convert everything to \( \sin x \) and \( \cos x \) first - this often reveals simplifications and cancellations that would have been hidden in the original form.

ANSWERS

Exercise 9.2

1. (i) \( \frac{2}{3a}(ax + b)^{3/2} + C \)
(ii) \( \frac{(ax + b)^{n+1}}{a(n+1)} + C, n \neq -1 \)

2. (i) \( -\frac{1}{3} \log |5 - 3x| + C \)
(ii) \( 2\sqrt{e^x} + C \)

3. (i) \( -\frac{1}{3} \tan(7 - 3x) + C \)
(ii) \( \frac{1}{2}(e^{2x} - e^{-2x} + 4x) + C \)

4. (i) \( \frac{x^3}{6} - \frac{x^2}{8} + \frac{x}{8} - \frac{1}{16} \log |2x + 1| + C \)
(ii) \( \frac{x^{3a+1}}{3a+1} + \frac{a^{3x}}{3 \log a} + C \)

5. (i) \( \frac{1}{9}[(2x+3)^{3/2} + (2x)^{3/2}] + C \)
(ii) \( \log |x + 1| + \frac{1}{x+1} + C \)

6. (i) \( \frac{1}{6}(2x-1)^{3/2} + \frac{3}{2}\sqrt{2x-1} + C \)
(ii) \( \frac{2}{45}(3x-2)^{5/2} + \frac{4}{27}(3x-2)^{3/2} + C \)

7. (i) \( \frac{1}{4}(2x + \sin 2x) + C \)
(ii) \( \frac{1}{32}(12x - 8 \sin 2x + \sin 4x) + C \)

8. (i) \( \frac{1}{8}(4x - \sin(4x + 10)) + C \)
(ii) \( \frac{1}{64}(24x + 8 \sin 4x + \sin 8x) + C \)

9. (i) \( 2\sqrt{2} \sin \frac{x}{2} + C \)
(ii) \( 2 \left( \sin \frac{x}{2} + \cos \frac{x}{2} \right) + C \)