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Math Notes For Class 12 Chapter 3 Ex 3.2 Solved & Quiz
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Math Notes For Class 12 Chapter 3 Ex 3.2 Solved & Quiz
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 ($a > 0$). |
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: $$\ln|\ln(x)| + C$$