Class 12 FSc Math Chapter 3 Integration Formulas & Solved Board Questions PDF | BISE Punjab & FBISE Study Guide

Theoretical Foundations & Core Theorems

Integration is one of the dual pillars of Calculus, serving as the inverse operation of differentiation. In FSc Part 2 Mathematics (Chapter 3), integration is introduced through two primary paradigms: the anti-derivative approach (Indefinite Integration) and the limit of a Riemann sum (Definite Integration).

1. The Concept of Anti-Derivatives and Indefinite Integrals

If $F(x)$ is a differentiable function such that its derivative with respect to $x$ yields $f(x)$, i.e.,

$$\frac{d}{dx}[F(x)] = f(x)$$

then $F(x)$ is termed an anti-derivative or primitive of $f(x)$. The process of finding $F(x)$ from $f(x)$ is called integration, denoted by the integral sign $\int$:

$$\int f(x) \, dx = F(x) + C$$

where $f(x)$ is the integrand, $dx$ is the differential element indicating integration with respect to $x$, and $C$ is the arbitrary constant of integration. The constant $C$ accounts for any constant term lost during differentiation, since $\frac{d}{dx}(C) = 0$. Geometrically, an indefinite integral represents a family of parallel curves.

2. The Fundamental Theorem of Calculus (FTC)

The Fundamental Theorem of Calculus bridges differential calculus and integral calculus, consisting of two main parts:

• First Fundamental Theorem: If $f(x)$ is continuous on a closed interval $[a, b]$ and $F(x) = \int_{a}^{x} f(t) \, dt$, then $F(x)$ is continuous on $[a, b]$, differentiable on $(a, b)$, and its derivative is: $$\frac{d}{dx} \left[ \int_{a}^{x} f(t) \, dt \right] = f(x)$$
• Second Fundamental Theorem (Evaluation Theorem): If $f(x)$ is continuous on $[a, b]$ and $F(x)$ is any anti-derivative of $f(x)$ such that $F'(x) = f(x)$, then the definite integral is evaluated as: $$\int_{a}^{b} f(x) \, dx = [F(x)]_{a}^{b} = F(b) - F(a)$$

3. The Differential Notation ($dx$) and Geometric Significance

The differential symbol $dx$ is not merely decorative; it represents an infinitesimally small increment along the $x$-axis. In definite integration, $\int_{a}^{b} f(x) \, dx$ represents the net signed area bounded by the curve $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$. Formally, this area is defined via the limit of a Riemann Sum:

$$\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$$

In physics and engineering applications taught across Pakistani universities (such as PU, NUST, and UOS), integration is vital for calculating work done by a variable force, centroids, fluid pressures, and solving dynamic differential equations.

Formula Summary & Quick Reference

Below is an exhaustive summary table of standard integration rules, techniques, and formulas required for BISE Board Examinations and University Entry Tests (ECAT/NUST):

Category	Formula / Identity	Conditions / Context
Power Rule	$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$	Valid for all $n \neq -1$
Power Rule (Composite)	$\int [f(x)]^n f'(x) \, dx = \frac{[f(x)]^{n+1}}{n+1} + C$	$n \neq -1$
Logarithmic Rule	$\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$	Denominator $f(x) \neq 0$
Exponential Functions	$\int e^{ax} \, dx = \frac{e^{ax}}{a} + C$ $\int a^x \, dx = \frac{a^x}{\ln a} + C$	$a > 0, a \neq 1$
Special Exponential Form	$\int e^{ax} [a f(x) + f'(x)] \, dx = e^{ax} f(x) + C$	Extremely important for ECAT & Board Exams
Trigonometric Integrals	$\int \sin(ax) \, dx = -\frac{\cos(ax)}{a} + C$ $\int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C$ $\int \sec^2(x) \, dx = \tan x + C$ $\int \csc^2(x) \, dx = -\cot x + C$ $\int \sec x \tan x \, dx = \sec x + C$ $\int \csc x \cot x \, dx = -\csc x + C$	Standard trigonometric anti-derivatives
Logarithmic Trigonometric Forms	$\int \tan x \, dx = \ln|\sec x| + C = -\ln|\cos x| + C$ $\int \cot x \, dx = \ln|\sin x| + C$ $\int \sec x \, dx = \ln|\sec x + \tan x| + C$ $\int \csc x \, dx = \ln|\csc x - \cot x| + C$	Derived via substitution
Inverse Trigonometric Forms	$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + C$ $\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$ $\int \frac{1}{x\sqrt{x^2 - a^2}} \, dx = \frac{1}{a} \sec^{-1}\left(\frac{x}{a}\right) + C$	Useful for trigonometric substitutions
Trigonometric Substitutions	For $\sqrt{a^2 - x^2} \implies$ Let $x = a \sin \theta$ For $\sqrt{a^2 + x^2} \implies$ Let $x = a \tan \theta$ For $\sqrt{x^2 - a^2} \implies$ Let $x = a \sec \theta$	Algebraic simplification to trigonometric identities
Integration by Parts	$\int u \, dv = u v - \int v \, du$	Selection order: LIATE (Logarithmic, Inverse trigo, Algebraic, Trigonometric, Exponential)
Definite Integral King's Property	$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx$	Simplifies symmetrical definite integrals

Step-by-Step Solved Board Exam Questions

Problem 1: Integration by Parts (Repeated Technique)

Question (BISE Lahore / FBISE Long Question): Evaluate the indefinite integral:

$$I = \int e^{ax} \sin(bx) \, dx$$

Solution:

Step 1: Apply Integration by Parts for the first time.

Using the LIATE rule, choose the first function $u = \sin(bx)$ and the second function $dv = e^{ax} dx$.

Calculate differentials and integrals:

$$du = b \cos(bx) \, dx$$ $$v = \int e^{ax} \, dx = \frac{e^{ax}}{a}$$

Applying the Integration by Parts formula $\int u \, dv = u v - \int v \, du$:

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

Step 2: Apply Integration by Parts a second time.

Let $I_1 = \int e^{ax} \cos(bx) \, dx$. Choose $u = \cos(bx)$ and $dv = e^{ax} dx$.

$$du = -b \sin(bx) \, dx, \quad v = \frac{e^{ax}}{a}$$ $$I_1 = \cos(bx) \cdot \frac{e^{ax}}{a} - \int \frac{e^{ax}}{a} (-b \sin(bx)) \, dx$$ $$I_1 = \frac{e^{ax} \cos(bx)}{a} + \frac{b}{a} \int e^{ax} \sin(bx) \, dx$$

Notice that $\int e^{ax} \sin(bx) \, dx = I$. Therefore:

$$I_1 = \frac{e^{ax} \cos(bx)}{a} + \frac{b}{a} I$$

Step 3: Substitute $I_1$ back into Equation 1 and solve for $I$.

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

Move the term containing $I$ to the left side:

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

Canceling $a^2$ from both denominators and dividing by $(a^2 + b^2)$:

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

Final Answer: $\int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} [a \sin(bx) - b \cos(bx)] + C$

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Problem 2: Integration via Trigonometric Substitution

Question (BISE Rawalpindi / Gujranwala Past Paper): Evaluate the standard integral:

$$I = \int \sqrt{a^2 - x^2} \, dx$$

Solution:

Step 1: Introduce trigonometric substitution.

Let $x = a \sin \theta \implies \sin \theta = \frac{x}{a} \implies \theta = \sin^{-1}\left(\frac{x}{a}\right)$.

Differentiating both sides gives: $dx = a \cos \theta \, d\theta$.

Step 2: Substitute into the integral.

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

Step 3: Use half-angle trigonometric identities.

Recall that $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$.

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

Step 4: Convert back to variable $x$.

Using the double-angle identity: $\sin(2\theta) = 2 \sin \theta \cos \theta$.

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

Substitute $\theta = \sin^{-1}\left(\frac{x}{a}\right)$, $\sin \theta = \frac{x}{a}$, and $\cos \theta = \frac{\sqrt{a^2 - x^2}}{a}$:

$$I = \frac{a^2}{2} \left[ \sin^{-1}\left(\frac{x}{a}\right) + \left(\frac{x}{a}\right)\left(\frac{\sqrt{a^2 - x^2}}{a}\right) \right] + C$$ $$I = \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + \frac{a^2}{2} \cdot \frac{x \sqrt{a^2 - x^2}}{a^2} + C$$ $$I = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$$

Final Answer: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$

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Problem 3: Definite Integral Property Proof

Question (FBISE / BISE Multan Long Question): Prove that:

$$\int_{0}^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} \, dx = \frac{\pi}{4}$$

Solution:

Step 1: Set up the integral.

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

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

Here, $a = \frac{\pi}{2}$. Replacing $x$ with $\left(\frac{\pi}{2} - x\right)$:

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

Step 3: Simplify using co-function identities.

Since $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\frac{\pi}{2} - x\right) = \sin x$:

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

Step 4: Add Equation 1 and Equation 2.

$$I + I = \int_{0}^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} \, dx + \int_{0}^{\pi/2} \frac{\cos^n x}{\sin^n x + \cos^n x} \, dx$$ $$2I = \int_{0}^{\pi/2} \left( \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x} \right) \, dx$$ $$2I = \int_{0}^{\pi/2} 1 \, dx$$ $$2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$ $$I = \frac{\pi}{4}$$

Hence Proved.

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Problem 4: Solving a First-Order Differential Equation

Question (BISE Sargodha / PU Entry Test): Find the general solution of the differential equation:

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

Solution:

Step 1: Separate the variables.

Group all terms containing $y$ with $dy$ and all terms containing $x$ with $dx$:

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

Step 2: Integrate both sides.

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

Step 3: Apply standard inverse trigonometric integration formulas.

Recall that $\int \frac{1}{1 + z^2} \, dz = \tan^{-1}(z) + C$.

$$\tan^{-1}(y) = \tan^{-1}(x) + C$$

Step 4: Explicitly express $y$ in terms of $x$ (Optional Alternative Form).

$$\tan^{-1}(y) - \tan^{-1}(x) = C$$

Taking tangent on both sides and using the subtraction formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:

$$\tan(\tan^{-1}(y) - \tan^{-1}(x)) = \tan(C)$$ $$\frac{y - x}{1 + xy} = C_1 \quad (\text{where } C_1 = \tan(C))$$ $$y - x = C_1 (1 + xy)$$

Final Answer: $\tan^{-1}(y) = \tan^{-1}(x) + C$ or $y - x = C_1 (1 + xy)$.

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

1. Why do we add a constant of integration $C$ for indefinite integrals but omit it for definite integrals?

In indefinite integration, we are finding a family of general functions whose derivative matches the integrand. Because the derivative of any constant $C$ is zero ($\frac{d}{dx}(C) = 0$), infinitely many functions differ by a constant but share the exact same derivative. Hence, $C$ is mandatory. In definite integration, we compute a specific net area between two limits $a$ and $b$:

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

The constant $C$ cancels out algebraically during evaluation, making its explicit inclusion unnecessary.

2. How do I determine which function to select as $u(x)$ and $v(x)$ during Integration by Parts?

To choose $u(x)$ effectively, follow the LIATE priority rule (from top priority for $u$ to lowest):

1. L: Logarithmic functions (e.g., $\ln x$)
2. I: Inverse Trigonometric functions (e.g., $\tan^{-1} x, \sin^{-1} x$)
3. A: Algebraic functions (e.g., $x^2, 3x, \sqrt{x}$)
4. T: Trigonometric functions (e.g., $\sin x, \cos x$)
5. E: Exponential functions (e.g., $e^x, 2^x$)

Select whichever function appears higher on the LIATE list as $u(x)$ (the function to be differentiated). The remaining factor becomes $dv$ (the function to be integrated).

3. What is the difference between a General Solution and a Particular Solution of a Differential Equation?

A General Solution of an $n$-th order differential equation contains $n$ arbitrary constants ($C_1, C_2, \dots$) and represents the entire family of curves satisfying the equation. A Particular Solution is obtained from the general solution by assigning specific numerical values to these arbitrary constants using given boundary conditions or initial conditions (e.g., $y(0) = 1$). A particular solution represents one single specific curve.