Math Notes For Class 12 Chapter 3 Ex 3.2 Solved & Quiz
What are the Core Formulas and Definitions for this Exercise?
Integration is the inverse process of differentiation. In Exercise 3.2, we focus on fundamental integration rules, power rule for composite functions, exponential/logarithmic rules, and basic method of substitution (variable transformation).
| Formula Name | Mathematical Expression | Applicability / Condition |
|---|---|---|
| Power Rule for Integrals | $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ | For any real number $n \neq -1$. |
| Extended Function Power Rule | $$\int [f(x)]^n f'(x) \, dx = \frac{[f(x)]^{n+1}}{n+1} + C$$ | When integrand contains function raised to power and its derivative ($n \neq -1$). |
| Logarithmic Integration Rule | $$\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$$ | When numerator is the exact derivative of denominator. |
| Exponential Rule | $$\int e^{kx} \, dx = \frac{e^{kx}}{k} + C$$ | Linear exponent $kx$ where $k \neq 0$. |
| Inverse Tangent Standard Form | $$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$$ | Quadratic expression in denominator ($a > 0$). |
| Inverse Sine Standard Form | $$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C$$ | Square root quadratic form ($a > 0$). |
How to Solve All Exercise Questions Step-by-Step?
Question 1 (Part i): Evaluate the indefinite integral: $\int (3x^2 - 2x + 1) \, dx$
Solution:
Apply the linearity property of integration to integrate term by term:
$$\int (3x^2 - 2x + 1) \, dx = 3 \int x^2 \, dx - 2 \int x^1 \, dx + \int 1 \, dx$$
Apply the basic power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1}$:
$$= 3 \left( \frac{x^{2+1}}{2+1} \right) - 2 \left( \frac{x^{1+1}}{1+1} \right) + x + C$$
$$= 3 \left( \frac{x^3}{3} \right) - 2 \left( \frac{x^2}{2} \right) + x + C$$
Simplify the coefficients:
$$= x^3 - x^2 + x + C$$
Answer: $$x^3 - x^2 + x + C$$
Question 1 (Part ii): Evaluate the indefinite integral: $\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) \, dx$
Solution:
Rewrite the radicals using fractional exponents:
$$\int \left(x^{1/2} + x^{-1/2}\right) \, dx = \int x^{1/2} \, dx + \int x^{-1/2} \, dx$$
Apply the power rule to both terms:
$$= \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + \frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C$$
$$= \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C$$
Invert the fractional denominators:
$$= \frac{2}{3} x^{3/2} + 2 \sqrt{x} + C$$
Answer: $$\frac{2}{3} x^{3/2} + 2 \sqrt{x} + C$$
Question 2 (Part i): Evaluate using substitution: $\int x \sqrt{x^2 + 5} \, dx$
Solution:
Let $u = x^2 + 5$. Differentiate both sides with respect to $x$:
$$\frac{du}{dx} = 2x \implies du = 2x \, dx \implies x \, dx = \frac{du}{2}$$
Substitute $u$ and $x \, dx$ into the original integral:
$$\int \sqrt{x^2 + 5} \cdot (x \, dx) = \int u^{1/2} \cdot \frac{du}{2} = \frac{1}{2} \int u^{1/2} \, du$$
Integrate using the power rule:
$$= \frac{1}{2} \left[ \frac{u^{1/2 + 1}}{\frac{1}{2} + 1} \right] + C = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right] + C = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C$$
Substitute back $u = x^2 + 5$:
$$= \frac{1}{3} (x^2 + 5)^{3/2} + C$$
Answer: $$\frac{1}{3} (x^2 + 5)^{3/2} + C$$
Question 2 (Part ii): Evaluate: $\int \frac{2x + 3}{x^2 + 3x + 7} \, dx$
Solution:
Examine the denominator: let $g(x) = x^2 + 3x + 7$.
Calculate its derivative:
$$g'(x) = \frac{d}{dx}(x^2 + 3x + 7) = 2x + 3$$
Notice that the numerator is equal to $g'(x)$.
Using the logarithmic rule $\int \frac{g'(x)}{g(x)} \, dx = \ln|g(x)| + C$:
$$\int \frac{2x + 3}{x^2 + 3x + 7} \, dx = \ln|x^2 + 3x + 7| + C$$
Answer: $$\ln|x^2 + 3x + 7| + C$$
Question 3 (Part i): Evaluate: $\int \cos(5x + 2) \, dx$
Solution:
Let $u = 5x + 2$. Differentiating yields:
$$du = 5 \, dx \implies dx = \frac{du}{5}$$
Substitute $u$ and $dx$ into the integral:
$$\int \cos(5x + 2) \, dx = \int \cos(u) \cdot \frac{du}{5} = \frac{1}{5} \int \cos(u) \, du$$
Since $\int \cos(u) \, du = \sin(u) + C$:
$$= \frac{1}{5} \sin(u) + C = \frac{1}{5} \sin(5x + 2) + C$$
Answer: $$\frac{1}{5} \sin(5x + 2) + C$$
Question 3 (Part ii): Evaluate: $\int \sec^2(3x) \tan(3x) \, dx$
Solution:
Let $u = \tan(3x)$. Differentiating yields:
$$du = 3 \sec^2(3x) \, dx \implies \sec^2(3x) \, dx = \frac{du}{3}$$
Substitute into integral:
$$\int \tan(3x) \left[\sec^2(3x) \, dx\right] = \int u \cdot \frac{du}{3} = \frac{1}{3} \int u \, du$$
Apply power rule:
$$= \frac{1}{3} \cdot \frac{u^2}{2} + C = \frac{1}{6} u^2 + C$$
Substitute back $u = \tan(3x)$:
$$= \frac{1}{6} \tan^2(3x) + C$$
Answer: $$\frac{1}{6} \tan^2(3x) + C$$
Question 4 (Part i): Evaluate: $\int e^{4x - 1} \, dx$
Solution:
Let $u = 4x - 1 \implies du = 4 \, dx \implies dx = \frac{du}{4}$.
Substitute into the integral:
$$\int e^{4x - 1} \, dx = \int e^u \cdot \frac{du}{4} = \frac{1}{4} \int e^u \, du$$
Since $\int e^u \, du = e^u + C$:
$$= \frac{1}{4} e^u + C = \frac{1}{4} e^{4x - 1} + C$$
Answer: $$\frac{1}{4} e^{4x - 1} + C$$
Question 4 (Part ii): Evaluate: $\int \frac{1}{x \ln(x)} \, dx$
Solution:
Rewrite the integrand as:
$$\int \frac{\frac{1}{x}}{\ln(x)} \, dx$$
Let $u = \ln(x)$. Then $du = \frac{1}{x} \, dx$.
Substitute $u$ and $du$:
$$\int \frac{1}{u} \, du = \ln|u| + C$$
Substitute back $u = \ln(x)$:
$$= \ln|\ln(x)| + C$$
Answer: $$\ln|\ln(x)| + C$$
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