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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
Indefinite integration (or anti-differentiation) is the reverse operation of differentiation. In maths class 12, mastering integration requires understanding fundamental differential properties, algebraic simplification, and variable substitution techniques.
1. Fundamental Theorem of Anti-derivatives:
If $F(x)$ is a function such that $\frac{d}{dx}[F(x)] = f(x)$, then the indefinite integral of $f(x)$ with respect to $x$ is defined as:
$$\int f(x) dx = F(x) + C$$
where $C$ is the constant of integration, representing an infinite family of parallel curves.
2. Method of Substitution ($u$-substitution):
When an integrand contains both a function $g(x)$ and its derivative $g'(x)$, we set $u = g(x)$, yielding $du = g'(x)dx$. The integral simplifies to:
$$\int f(g(x)) g'(x) dx = \int f(u) du$$
Below is a quick reference table of essential integration formulas for Exercise 3.2:
| Rule / Category | Standard Formula | Generalized Composite Formula |
|---|---|---|
| Power Rule ($n \neq -1$) | $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ | $$\int [f(x)]^n f'(x) dx = \frac{[f(x)]^{n+1}}{n+1} + C$$ |
| Logarithmic Rule | $$\int \frac{1}{x} dx = \ln|x| + C$$ | $$\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C$$ |
| Exponential Rule | $$\int e^x dx = e^x + C$$ | $$\int e^{ax} dx = \frac{e^{ax}}{a} + C$$ |
| Base-$a$ Exponential | $$\int a^x dx = \frac{a^x}{\ln a} + C$$ | $$\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C$$ |
| Trigonometric Integrals | $$\int \cos x dx = \sin x + C$$ | $$\int \sin x dx = -\cos x + C$$ |
Question 1 (Part i): Evaluate the indefinite integral: $\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx$
Solution:
Express square roots as fractional exponents:
$$\int \left(x^{1/2} + x^{-1/2}\right) dx$$
Apply the linearity property of integration to separate terms:
$$= \int x^{1/2} dx + \int x^{-1/2} dx$$
Apply the Power Rule of Integration ($\int x^n dx = \frac{x^{n+1}}{n+1} + C$):
$$= \frac{x^{1/2 + 1}}{\frac{1}{2} + 1} + \frac{x^{-1/2 + 1}}{-\frac{1}{2} + 1} + C$$
$$= \frac{x^{3/2}}{\frac{3}{2}} + \frac{x^{1/2}}{\frac{1}{2}} + C$$
Simplify fractions:
$$= \frac{2}{3}x^{3/2} + 2x^{1/2} + C = \frac{2}{3}x^{3/2} + 2\sqrt{x} + C$$
Answer: $$\frac{2}{3}x^{3/2} + 2\sqrt{x} + C$$
Question 1 (Part ii): Evaluate the integral: $\int \frac{(1 - \sqrt{x})^2}{\sqrt{x}} dx$
Solution:
Expand the numerator using algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$:
$$(1 - \sqrt{x})^2 = 1 - 2\sqrt{x} + (\sqrt{x})^2 = 1 - 2\sqrt{x} + x$$
Divide each term in the numerator by $\sqrt{x} = x^{1/2}$:
$$\frac{1 - 2x^{1/2} + x}{x^{1/2}} = \frac{1}{x^{1/2}} - \frac{2x^{1/2}}{x^{1/2}} + \frac{x}{x^{1/2}} = x^{-1/2} - 2 + x^{1/2}$$
Integrate term-by-term:
$$\int \left(x^{-1/2} - 2 + x^{1/2}\right) dx = \int x^{-1/2} dx - 2 \int 1 dx + \int x^{1/2} dx$$
$$= \frac{x^{1/2}}{\frac{1}{2}} - 2x + \frac{x^{3/2}}{\frac{3}{2}} + C$$
$$= 2\sqrt{x} - 2x + \frac{2}{3}x^{3/2} + C$$
Answer: $$2\sqrt{x} - 2x + \frac{2}{3}x^{3/2} + C$$
Question 2 (Part i): Evaluate the trigonometric integral: $\int \sin^2 x \, dx$
Solution:
Recall the half-angle trigonometric identity: $\sin^2 x = \frac{1 - \cos(2x)}{2}$.
Substitute this into the integrand:
$$\int \sin^2 x \, dx = \int \frac{1 - \cos(2x)}{2} dx$$
Separate into two distinct integrals:
$$= \frac{1}{2} \int 1 dx - \frac{1}{2} \int \cos(2x) dx$$
Integrate each component:
$$\int 1 dx = x, \quad \int \cos(2x) dx = \frac{\sin(2x)}{2}$$
Combining results:
$$= \frac{1}{2} x - \frac{1}{2} \left(\frac{\sin(2x)}{2}\right) + C = \frac{1}{2}x - \frac{1}{4}\sin(2x) + C$$
Answer: $$\frac{1}{2}x - \frac{1}{4}\sin(2x) + C$$
Question 2 (Part ii): Evaluate the integral: $\int \frac{\cos x}{\sqrt{\sin x}} \, dx$
Solution:
Use the substitution method. Let $u = \sin x$.
Differentiating $u$ with respect to $x$ gives:
$$\frac{du}{dx} = \cos x \implies du = \cos x \, dx$$
Substitute $u$ and $du$ into the original integral:
$$\int \frac{\cos x \, dx}{\sqrt{\sin x}} = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} du$$
Apply the Power Rule:
$$= \frac{u^{-1/2 + 1}}{-\frac{1}{2} + 1} + C = \frac{u^{1/2}}{\frac{1}{2}} + C = 2\sqrt{u} + C$$
Substitute back $u = \sin x$:
$$= 2\sqrt{\sin x} + C$$
Answer: $$2\sqrt{\sin x} + C$$
Question 3 (Part i): Evaluate using variable substitution: $\int x \sqrt{x^2 + 5} \, dx$
Solution:
Let $u = x^2 + 5$.
Differentiating yields $\frac{du}{dx} = 2x \implies du = 2x \, dx \implies x \, dx = \frac{du}{2}$.
Substitute into the integral:
$$\int \sqrt{x^2 + 5} \cdot (x dx) = \int \sqrt{u} \cdot \left(\frac{du}{2}\right) = \frac{1}{2} \int u^{1/2} du$$
Apply Power Rule:
$$= \frac{1}{2} \left( \frac{u^{3/2}}{3/2} \right) + C = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C$$
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 3 (Part ii): Evaluate the integral: $\int \frac{e^x}{1 + e^x} \, dx$
Solution:
Observe that the numerator is the exact derivative of the denominator:
$$\frac{d}{dx}(1 + e^x) = e^x$$
Let $u = 1 + e^x$, then $du = e^x dx$.
Using the logarithmic derivative rule $\int \frac{g'(x)}{g(x)} dx = \ln|g(x)| + C$:
$$\int \frac{e^x dx}{1 + e^x} = \int \frac{du}{u} = \ln|u| + C$$
Substitute $u = 1 + e^x$ back (since $1 + e^x > 0$ for all real $x$, absolute value bars can be changed to parentheses):
$$= \ln(1 + e^x) + C$$
Answer: $$\ln(1 + e^x) + C$$
Question 4 (Part i): Evaluate the logarithmic integrand: $\int \frac{1}{x \ln x} \, dx$
Solution:
Rewrite the integrand to highlight the ratio of a function and its derivative:
$$\int \frac{1}{x \ln x} dx = \int \frac{\frac{1}{x}}{\ln x} dx$$
Let $u = \ln x$. Differential $du = \frac{1}{x} dx$.
Substitute into the integral:
$$\int \frac{du}{u} = \ln|u| + C$$
Substitute back $u = \ln x$:
$$= \ln|\ln x| + C$$
Answer: $$\ln|\ln x| + C$$
Question 4 (Part ii): Evaluate the exponential integral: $\int a^{2x} \, dx \quad (a > 0, a \neq 1)$
Solution:
Recall the formula for integrating exponential functions of base $a$: $\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C$.
Alternatively, substitute $u = 2x \implies du = 2dx \implies dx = \frac{du}{2}$:
$$\int a^{2x} dx = \int a^u \left(\frac{du}{2}\right) = \frac{1}{2} \int a^u du = \frac{1}{2} \cdot \frac{a^u}{\ln a} + C$$
Substitute $u = 2x$ back:
$$= \frac{a^{2x}}{2 \ln a} + C$$
Answer: $$\frac{a^{2x}}{2 \ln a} + C$$
Question 5 (Part i): Evaluate: $\int \frac{x+2}{x^2 + 4x + 7} \, dx$
Solution:
Let $u = x^2 + 4x + 7$.
Compute derivative:
$$\frac{du}{dx} = 2x + 4 = 2(x + 2) \implies (x + 2) dx = \frac{du}{2}$$
Substitute this into the expression:
$$\int \frac{x+2}{x^2 + 4x + 7} dx = \int \frac{\frac{du}{2}}{u} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| + C$$
Substitute $u = x^2 + 4x + 7$ back (note $x^2 + 4x + 7 = (x+2)^2 + 3 > 0$ for all real $x$):
$$= \frac{1}{2} \ln(x^2 + 4x + 7) + C$$
Answer: $$\frac{1}{2} \ln(x^2 + 4x + 7) + C$$
Question 5 (Part ii): Evaluate: $\int \frac{\sec^2 x}{\tan x} \, dx$
Solution:
Notice that $\frac{d}{dx}(\tan x) = \sec^2 x$.
Let $u = \tan x \implies du = \sec^2 x \, dx$.
Substitute into the integrand:
$$\int \frac{\sec^2 x}{\tan x} dx = \int \frac{du}{u} = \ln|u| + C$$
Substitute $u = \tan x$ back:
$$= \ln|\tan x| + C$$
Answer: $$\ln|\tan x| + C$$
Question 6: Find the function $f(x)$ if its derivative is given by $f'(x) = 3x^2 - 4x + 5$ and $f(1) = 6$.
Solution:
Integrate $f'(x)$ to find the general expression for $f(x)$:
$$f(x) = \int f'(x) dx = \int (3x^2 - 4x + 5) dx$$
Apply power rule term-by-term:
$$f(x) = 3 \left(\frac{x^3}{3}\right) - 4 \left(\frac{x^2}{2}\right) + 5x + C$$
$$f(x) = x^3 - 2x^2 + 5x + C$$
Use the initial condition $f(1) = 6$ to determine constant $C$:
$$6 = (1)^3 - 2(1)^2 + 5(1) + C$$
$$6 = 1 - 2 + 5 + C \implies 6 = 4 + C \implies C = 2$$
Substitute $C = 2$ back into the function:
$$f(x) = x^3 - 2x^2 + 5x + 2$$
Answer: $$f(x) = x^3 - 2x^2 + 5x + 2$$
Q1: What is the antiderivative of $x^{-1}$?
Q2: What is $\int e^{3x} dx$?
Q3: What is the result of $\int \frac{f'(x)}{f(x)} dx$?
Q4: Evaluate $\int \cos(5x) dx$:
Q5: Value of $\int 2^x dx$ is:
Q6: What trigonometric identity is used to integrate $\cos^2 x$?