Class 12 FSc Math Chapter 3 Integration Formulas & Solved Important Questions PDF Guide | FBISE & Punjab Boards

Theoretical Foundations & Core Theorems

Integration forms the bedrock of modern mathematical analysis and calculus, serving as the inverse process of differentiation. In FSc Part II Mathematics (Chapter 3), integration is approached from two primary perspectives: anti-differentiation (indefinite integrals) and the limiting process of Riemann Sums (definite integrals).

1. The Fundamental Theorem of Calculus (FTC)

The bridge between differentiation and integration is formally established by the Fundamental Theorem of Calculus. Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$.

First Fundamental Theorem of Calculus: If $F(x)$ is defined by $F(x) = \int_{a}^{x} f(t) \, dt$ for all $x \in [a, b]$, then $F(x)$ is continuous on $[a, b]$, differentiable on the open interval $(a, b)$, and its derivative is given by:

$$F'(x) = \frac{d}{dx} \left[ \int_{a}^{x} f(t) \, dt \right] = f(x)$$

Second Fundamental Theorem of Calculus: If $F(x)$ is any antiderivative of $f(x)$ such that $F'(x) = f(x)$, then the definite integral of $f(x)$ over $[a, b]$ is evaluated as:

$$\int_{a}^{b} f(x) \, dx = [F(x)]_{a}^{b} = F(b) - F(a)$$

2. Theoretical Foundations of Integration Techniques

• Integration by Substitution (Change of Variable): Derived directly from the differentiation Chain Rule. If $u = g(x)$ is a differentiable function whose range is an interval $I$, and $f$ is continuous on $I$, then: $$\int f(g(x)) g'(x) \, dx = \int f(u) \, du$$
• Integration by Parts: Derived from the Product Rule of differentiation $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$. Integrating both sides with respect to $x$ yields the standard formula: $$\int u \, dv = u v - \int v \, du \quad \text{or} \quad \int f(x) g(x) \, dx = f(x) \int g(x) \, dx - \int \left( f'(x) \int g(x) \, dx \right) dx$$

3. Real-World Significance & Engineering Applications

In physical sciences and engineering, integration allows for the transition from instantaneous rates of change to cumulative quantities. Applications include calculating the work done by a variable force ($W = \int F \, dx$), determining center of mass and moments of inertia, evaluating signal energy in electrical circuits, computing probability density distributions in statistics, and modeling dynamic systems through differential equations.

Formula Summary & Quick Reference

Below is a curated reference table containing the core integration formulas essential for FSc board examinations and Pakistani engineering entry tests (ECAT, NUST NET).

Category / Rule	Standard Integration Formula / Condition	Primary Application
Power Rule	$$\int [f(x)]^n f'(x) \, dx = \frac{[f(x)]^{n+1}}{n+1} + C \quad (n \neq -1)$$	Polynomials, algebraic expressions with internal derivative present.
Logarithmic Rule	$$\int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C$$	Rational functions where numerator is derivative of denominator ($n = -1$).
Exponential Functions	$$\int e^{ax} \, dx = \frac{e^{ax}}{a} + C, \quad \int a^x \, dx = \frac{a^x}{\ln a} + C$$	Population dynamics, radioactive decay, circuit discharge.
Special Exponential Form	$$\int e^{ax} \left[ a f(x) + f'(x) \right] \, dx = e^{ax} f(x) + C$$	Advanced Board/ECAT long questions involving product of $e^{ax}$ and trig/inverse trig.
Trigonometric Functions	$$\int \sin x \, dx = -\cos x + C, \quad \int \cos x \, dx = \sin x + C$$ $$\int \sec^2 x \, dx = \tan x + C, \quad \int \sec x \tan x \, dx = \sec x + C$$ $$\int \tan x \, dx = \ln|\sec x| + C, \quad \int \sec x \, dx = \ln|\sec x + \tan x| + C$$	Harmonic motion, wave mechanics, geometric transformations.
Inverse Trigonometric Forms	$$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C$$ $$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$$ $$\int \frac{1}{x\sqrt{x^2 - a^2}} \, dx = \frac{1}{a} \text{arcsec}\left(\frac{x}{a}\right) + C$$	Evaluations requiring trigonometric substitutions ($x = a\sin\theta, x=a\tan\theta$).
Definite Integral Symmetry Property	$$\int_{-a}^{a} f(x) \, dx = 0 \quad \text{if } f(-x) = -f(x) \text{ (Odd Function)}$$	Rapid simplification of definite integrals over symmetric intervals.

Step-by-Step Solved Board Exam Questions

Question 1 (Standard Long Question - Trigonometric Substitution)

Problem: Evaluate the standard integral: $$I = \int \sqrt{a^2 - x^2} \, dx$$

Solution:

Step 1: Apply the trigonometric substitution strategy. Let $x = a \sin \theta$.

Differentiating both sides with respect to $\theta$ gives:

$$dx = a \cos \theta \, d\theta$$

Also, from the substitution, $\sin \theta = \frac{x}{a} \implies \theta = \arcsin\left(\frac{x}{a}\right)$.

Step 2: Substitute $x$ and $dx$ into the integral $I$:

$$I = \int \sqrt{a^2 - a^2 \sin^2 \theta} \cdot (a \cos \theta) \, d\theta$$ $$I = \int \sqrt{a^2(1 - \sin^2 \theta)} \cdot a \cos \theta \, d\theta$$

Using the Pythagorean identity $1 - \sin^2 \theta = \cos^2 \theta$:

$$I = \int \sqrt{a^2 \cos^2 \theta} \cdot a \cos \theta \, d\theta = \int (a \cos \theta) \cdot (a \cos \theta) \, d\theta$$ $$I = a^2 \int \cos^2 \theta \, d\theta$$

Step 3: Use the half-angle identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$:

$$I = a^2 \int \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{a^2}{2} \int (1 + \cos(2\theta)) \, d\theta$$ $$I = \frac{a^2}{2} \left[ \theta + \frac{\sin(2\theta)}{2} \right] + C$$

Step 4: Apply the double-angle formula $\sin(2\theta) = 2 \sin \theta \cos \theta$:

$$I = \frac{a^2}{2} \left[ \theta + \frac{2 \sin \theta \cos \theta}{2} \right] + C = \frac{a^2}{2} \left[ \theta + \sin \theta \cos \theta \right] + C$$

Step 5: Convert back to the original variable $x$. Since $\sin \theta = \frac{x}{a}$, we construct a right triangle where $\cos \theta = \frac{\sqrt{a^2 - x^2}}{a}$:

$$I = \frac{a^2}{2} \left[ \arcsin\left(\frac{x}{a}\right) + \left(\frac{x}{a}\right)\left(\frac{\sqrt{a^2 - x^2}}{a}\right) \right] + C$$ $$I = \frac{a^2}{2} \arcsin\left(\frac{x}{a}\right) + \frac{a^2}{2} \cdot \frac{x \sqrt{a^2 - x^2}}{a^2} + C$$ $$I = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \arcsin\left(\frac{x}{a}\right) + C$$ ---

Question 2 (Cyclic Integration by Parts)

Problem: Evaluate the integral: $$I = \int e^{ax} \sin(bx) \, dx$$

Solution:

Step 1: Apply Integration by Parts using ILATE rule (Choose $u = \sin(bx)$ and $dv = e^{ax} dx$).

$$\begin{aligned} u &= \sin(bx) &\implies du &= b \cos(bx) \, dx \\ dv &= e^{ax} \, dx &\implies v &= \frac{e^{ax}}{a} \end{aligned}$$

Using the integration by parts formula $\int u \, dv = uv - \int v \, du$:

$$I = \sin(bx) \cdot \frac{e^{ax}}{a} - \int \frac{e^{ax}}{a} \cdot b \cos(bx) \, dx$$ $$I = \frac{e^{ax} \sin(bx)}{a} - \frac{b}{a} \int e^{ax} \cos(bx) \, dx \quad \text{--- (Equation 1)}$$

Step 2: Apply Integration by Parts a second time on $\int e^{ax} \cos(bx) \, dx$.

Let $u_1 = \cos(bx)$ and $dv_1 = e^{ax} dx$:

$$\begin{aligned} u_1 &= \cos(bx) &\implies du_1 &= -b \sin(bx) \, dx \\ dv_1 &= e^{ax} \, dx &\implies v_1 &= \frac{e^{ax}}{a} \end{aligned}$$ $$\int e^{ax} \cos(bx) \, dx = \frac{e^{ax} \cos(bx)}{a} - \int \frac{e^{ax}}{a} (-b \sin(bx)) \, dx$$ $$\int e^{ax} \cos(bx) \, dx = \frac{e^{ax} \cos(bx)}{a} + \frac{b}{a} \int e^{ax} \sin(bx) \, dx$$

Step 3: Substitute this expression back into Equation 1:

$$I = \frac{e^{ax} \sin(bx)}{a} - \frac{b}{a} \left[ \frac{e^{ax} \cos(bx)}{a} + \frac{b}{a} I \right]$$ $$I = \frac{e^{ax} \sin(bx)}{a} - \frac{b e^{ax} \cos(bx)}{a^2} - \frac{b^2}{a^2} I$$

Step 4: Collect terms containing $I$ on the left-hand side:

$$I + \frac{b^2}{a^2} I = \frac{e^{ax}}{a^2} \left[ a \sin(bx) - b \cos(bx) \right]$$ $$I \left( 1 + \frac{b^2}{a^2} \right) = \frac{e^{ax}}{a^2} \left[ a \sin(bx) - b \cos(bx) \right]$$ $$I \left( \frac{a^2 + b^2}{a^2} \right) = \frac{e^{ax}}{a^2} \left[ a \sin(bx) - b \cos(bx) \right]$$

Step 5: Multiply both sides by $\frac{a^2}{a^2 + b^2}$ and add constant of integration $C$:

$$I = \frac{e^{ax}}{a^2 + b^2} \left[ a \sin(bx) - b \cos(bx) \right] + C$$ ---

Question 3 (Definite Integral Property Application)

Problem: Evaluate the definite integral: $$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx$$

Solution:

Step 1: State the primary equation:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx \quad \text{--- (Equation 1)}$$

Step 2: Apply the Definite Integral Property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$. Here $a = \frac{\pi}{2}$:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)} \, dx$$

Step 3: Use co-function identities $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\frac{\pi}{2} - x\right) = \sin x$:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} \, dx \quad \text{--- (Equation 2)}$$

Step 4: Add Equation 1 and Equation 2:

$$I + I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx + \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \, dx$$ $$2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\sin x + \cos x}{\sin x + \cos x} \right) \, dx$$ $$2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx$$

Step 5: Integrate and apply limits:

$$2I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$ $$I = \frac{\pi}{4}$$ ---

Question 4 (First Order Differential Equation)

Problem: Solve the initial value differential equation: $$\frac{dy}{dx} + \frac{1+y^2}{1+x^2} = 0, \quad \text{given } y(0) = 1$$

Solution:

Step 1: Rearrange the equation to isolate terms:

$$\frac{dy}{dx} = -\frac{1+y^2}{1+x^2}$$

Step 2: Separate variables by placing all $y$-terms on the left side and $x$-terms on the right side:

$$\frac{1}{1+y^2} \, dy = -\frac{1}{1+x^2} \, dx$$

Step 3: Integrate both sides with respect to their corresponding variables:

$$\int \frac{1}{1+y^2} \, dy = -\int \frac{1}{1+x^2} \, dx$$ $$\arctan(y) = -\arctan(x) + C \quad \text{--- (General Solution)}$$

Step 4: Apply the initial condition $y(0) = 1$ (substitute $x = 0$ and $y = 1$) to find $C$:

$$\arctan(1) = -\arctan(0) + C$$ $$\frac{\pi}{4} = 0 + C \implies C = \frac{\pi}{4}$$

Step 5: Substitute $C = \frac{\pi}{4}$ back into the general solution:

$$\arctan(y) + \arctan(x) = \frac{\pi}{4}$$

Using the identity $\arctan(A) + \arctan(B) = \arctan\left(\frac{A+B}{1-AB}\right)$:

$$\arctan\left(\frac{y+x}{1-xy}\right) = \frac{\pi}{4}$$ $$\frac{y+x}{1-xy} = \tan\left(\frac{\pi}{4}\right) = 1$$ $$y + x = 1 - xy \implies x + y + xy = 1$$

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

Q1: How do I select the first function $u$ and second function $v$ in Integration by Parts?
Answer: Follow the traditional ILATE priority rule (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential). The function appearing earlier in ILATE should be chosen as the first function $u$ (to be differentiated), while the function appearing later should be chosen as $dv$ (to be integrated). For example, in $\int x \ln x \, dx$, $\ln x$ is Logarithmic (L) and $x$ is Algebraic (A), so set $u = \ln x$ and $dv = x \, dx$.

Q2: Why is the arbitrary constant $+C$ added in indefinite integrals, but omitted in definite integrals?
Answer: An indefinite integral represents a family of parallel curves differing by a vertical shift. Because the derivative of any constant $C$ is zero ($\frac{d}{dx}[F(x) + C] = f(x)$), we must add $+C$ to preserve completeness. In definite integrals, when calculating $[F(b) + C] - [F(a) + C]$, the constants cancel out: $(F(b) + C) - (F(a) + C) = F(b) - F(a)$, rendering $+C$ unnecessary.

Q3: How do I identify when to use Partial Fraction Decomposition for Integration in FSc Chapter 3?
Answer: Use Partial Fractions when integrating rational functions $P(x)/Q(x)$ where: 1. The denominator $Q(x)$ can be factored into linear or irreducible quadratic factors. 2. The fraction is proper (degree of numerator $P(x)$ is strictly less than degree of denominator $Q(x)$). If the fraction is improper, perform polynomial long division first until you obtain a polynomial plus a proper rational fraction, then apply partial fraction decomposition.