Integrals Class 12 NCERT: Concepts, Methods, and Solved Examples

Understanding Integration: The Anti-Derivative and Area Under Curve

Integration is fundamentally the inverse process of differentiation. If the derivative of a function $F(x)$ is $f(x)$, i.e., $F'(x) = f(x)$, then the indefinite integral of $f(x)$ with respect to $x$ is $F(x) + C$, written mathematically as $\int f(x) \, dx = F(x) + C$. Here, $C$ is the arbitrary constant of integration because the derivative of any constant is zero. Physically, indefinite integration gives a family of curves. Definite integration, denoted by $\int_{a}^{b} f(x) \, dx$, calculates the net signed area under the curve $y = f(x)$ from $x=a$ to $x=b$. By the Fundamental Theorem of Calculus, this is evaluated as $F(b) - F(a)$. Success in this chapter requires mastering three core methods: substitution (changing variables to simplify), partial fractions (breaking down rational functions), and integration by parts (integrating products of functions using the ILATE rule).

Core Definitions and Methods of Integration

Indefinite Integral

The process of finding the general anti-derivative of a function, containing an arbitrary constant C representing a family of curves.

Method of Substitution

A technique where a variable x is replaced by a new variable t = g(x) such that g'(x)dx becomes dt, simplifying the integrand into a standard form.

Integration by Parts

A method based on the product rule of differentiation, expressed as the integral of the product of two functions using the ILATE priority rule.

Fundamental Theorem of Calculus

The theorem connecting differentiation and integration, stating that if f is continuous on [a, b] and F is the anti-derivative, then the definite integral equals F(b) - F(a).

Step-by-Step Worked Examples

• Example 1: Integrate $\int x \cos(x) \, dx$ using Integration by Parts. Step 1: Identify parts using the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential). Here, $u = x$ (Algebraic) and $v = \cos(x)$ (Trigonometric). Step 2: Apply the integration by parts formula: $\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$. Step 3: Differentiate $u$: $u' = 1$. Integrate $v$: $\int \cos(x) \, dx = \sin(x)$. Step 4: Substitute these into the formula: $\int x \cos(x) \, dx = x \sin(x) - \int (1 \cdot \sin(x)) \, dx$. Step 5: Integrate the remaining term: $-\int \sin(x) \, dx = -(-\cos(x)) = \cos(x)$. Step 6: Combine terms and add the constant of integration C. Final answer: $x \sin(x) + \cos(x) + C$
• Example 2: Evaluate $\int_{0}^{1} \frac{2x}{1+x^2} \, dx$ using the substitution method. Step 1: Choose a substitution. Let $t = 1 + x^2$. Then, differentiating both sides gives $dt = 2x \, dx$. Step 2: Change the limits of integration. When $x = 0$, $t = 1 + 0^2 = 1$. When $x = 1$, $t = 1 + 1^2 = 2$. Step 3: Rewrite the integral in terms of $t$ with the new limits: $\int_{1}^{2} \frac{1}{t} \, dt$. Step 4: Integrate the function: $\int \frac{1}{t} \, dt = [\ln|t|]_{1}^{2}$. Step 5: Apply the limits: $\ln|2| - \ln|1| = \ln(2) - 0$. Final answer: $\ln(2)$

Avoid These Common Board Exam Traps

1. Forgetting the Constant of Integration ($C$): For indefinite integrals, always write $+ C$ at the end. Forgetting this is an automatic half-mark deduction in CBSE evaluation.
2. Not Updating Limits in Definite Integrals: When using substitution $t = g(x)$ for a definite integral, you must transform the lower and upper limits. If you keep the original limits of $x$, your final evaluation will be incorrect.
3. Incorrect ILATE Order: When performing integration by parts, choosing the wrong first function $u$ can make the integral impossible to solve. Always strictly follow the ILATE hierarchy (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential).

Practice Questions with Solutions

• Q: Find the integral of $\int e^x (\sin(x) + \cos(x)) \, dx$. A: Step 1: Recognize the standard form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$. Step 2: Identify $f(x)$ and $f'(x)$. Let $f(x) = \sin(x)$. Then, its derivative $f'(x) = \cos(x)$. Step 3: Since the integrand perfectly matches the standard form, apply the rule directly. Final answer: $e^x \sin(x) + C$
• Q: Evaluate $\int \frac{1}{x^2 - 16} \, dx$. A: Step 1: Recognize the standard special integral formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right| + C$. Step 2: Rewrite $16$ as $4^2$. Thus, $a = 4$. Step 3: Substitute $a = 4$ into the formula: $\frac{1}{2(4)} \ln \left| \frac{x-4}{x+4} \right| + C$. Final answer: $\frac{1}{8} \ln \left| \frac{x-4}{x+4} \right| + C$
• Q: Evaluate the definite integral $I = \int_{0}^{\pi/2} \frac{\sin^4(x)}{\sin^4(x) + \cos^4(x)} \, dx$. A: Step 1: Apply the property of definite integrals: $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. Here, replace $x$ with $\frac{\pi}{2} - x$. Step 2: Rewrite the integral: $I = \int_{0}^{\pi/2} \frac{\cos^4(x)}{\cos^4(x) + \sin^4(x)} \, dx$. Step 3: Add the original integral and the modified integral: $2I = \int_{0}^{\pi/2} \frac{\sin^4(x) + \cos^4(x)}{\sin^4(x) + \cos^4(x)} \, dx = \int_{0}^{\pi/2} 1 \, dx$. Step 4: Integrate and evaluate: $2I = [x]_{0}^{\pi/2} = \frac{\pi}{2}$. Therefore, $I = \frac{\pi}{4}$. Final answer: $\frac{\pi}{4}$
• Q: Evaluate $\int x \ln(x) \, dx$. A: Step 1: Use Integration by Parts. According to ILATE, let $u = \ln(x)$ (Logarithmic) and $v = x$ (Algebraic). Step 2: Differentiate $u$: $u' = \frac{1}{x}$. Integrate $v$: $\int x \, dx = \frac{x^2}{2}$. Step 3: Apply the integration by parts formula: $\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$. Step 4: Substitute the values: $\ln(x) \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx = \frac{x^2}{2} \ln(x) - \frac{1}{2} \int x \, dx$. Step 5: Solve the final integration: $-\frac{1}{2} \cdot \frac{x^2}{2} = -\frac{x^2}{4}$. Add constant $C$. Final answer: $\frac{x^2}{2} \ln(x) - \frac{x^2}{4} + C$

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