Class 12 FSc Math Chapter 3 Integration Formulas, Proofs & Solved Past Paper Long Questions PDF Study Guide | FBISE & Punjab Boards
Welcome to the ultimate academic module for Class 12 FSc Mathematics Chapter 3: Integration. Integration (often termed anti-differentiation or integral calculus) forms the backbone of modern analysis, differential equations, physics, and engineering mechanics. In the Pakistani educational system (both Intermediate Boards such as FBISE, BISE Lahore, BISE Rawalpindi, BISE Multan, BISE Faisalabad, and public university entrance tests like ECAT, NUST NET, FAST, and GIKI), Chapter 3 carries the highest weightage—often accounting for 25% to 30% of the total examination score.
This guide is engineered to bridge conceptual clarity with board-specific scoring criteria, providing precise derivations, rigorous mathematical proofs, and step-by-step problem-solving methods.
Integral calculus is fundamentally concerned with two complementary operations: finding the antiderivative of a function (Indefinite Integration) and calculating the net area under a curve over a given interval (Definite Integration).
Let $f(x)$ be a real-valued continuous function defined on an open interval $I$. A function $F(x)$ is defined as an antiderivative or primitive of $f(x)$ if its derivative equals $f(x)$ for all $x \in I$:
$$\frac{d}{dx}\left[F(x)\right] = f(x)$$The collection of all antiderivatives of $f(x)$ is represented by the indefinite integral:
$$\int f(x) \, dx = F(x) + c$$where $\int$ denotes the integral sign, $f(x)$ is the integrand, $dx$ signifies the variable of integration, and $c \in \mathbb{R}$ represents the arbitrary constant of integration. The constant $c$ accounts for the family of parallel curves whose instantaneous rates of change are identical.
Geometrically, the definite integral measures the net signed area bounded by the graph of $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$. Partitioning the interval $[a, b]$ into $n$ subintervals of equal width $\Delta x = \frac{b-a}{n}$, the Riemann Sum is defined as:
$$\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x = \int_{a}^{b} f(x) \, dx$$where $x_i^*$ is a sample point in the $i$-th subinterval $[x_{i-1}, x_i]$.
The bridge between differential calculus and integral calculus is governed by two vital theorems:
The following table serves as an exhaustive reference guide for board examinations and entrance tests. Master these basic and advanced integration formulas:
| Key Concept / Rule | Mathematical Formula / Standard Form | Board & Entry Test Application Context |
|---|---|---|
| Basic Power Rule | $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + c \quad (n \neq -1)$$ | Polynomial terms, radical expressions, power functions. |
| Generalized Power Rule | $$\int [f(x)]^n f'(x) \, dx = \frac{[f(x)]^{n+1}}{n+1} + c \quad (n \neq -1)$$ | Integrands containing a function raised to a power multiplied by its derivative. |
| Logarithmic Rule | $$\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + c$$ | Fractions where the numerator is the exact derivative of the denominator. |
| Exponential Rules | $$\int e^{kx} \, dx = \frac{e^{kx}}{k} + c, \quad \int a^{kx} \, dx = \frac{a^{kx}}{k \ln a} + c$$ | Natural and general exponential functions in growth models and physics. |
| Trigonometric Integrals |
$$\int \sin x \, dx = -\cos x + c, \quad \int \cos x \, dx = \sin x + c$$ $$\int \sec^2 x \, dx = \tan x + c, \quad \int \csc^2 x \, dx = -\cot x + c$$ |
Core identity transformations and periodic evaluations. |
| Logarithmic Trig Integrals |
$$\int \tan x \, dx = \ln|\sec x| + c, \quad \int \cot x \, dx = \ln|\sin x| + c$$ $$\int \sec x \, dx = \ln|\sec x + \tan x| + c, \quad \int \csc x \, dx = \ln|\csc x - \cot x| + c$$ |
Advanced trigonometric integration and substitutions. |
| Inverse Trig Forms |
$$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + c$$ $$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c$$ $$\int \frac{1}{x\sqrt{x^2 - a^2}} \, dx = \frac{1}{a} \sec^{-1}\left(\frac{x}{a}\right) + c$$ |
Algebraic integrand reductions using inverse trigonometric substitutions ($x = a\sin\theta, a\tan\theta, a\sec\theta$). |
| Special Exponential Theorem | $$\int e^{ax} \left[ a f(x) + f'(x) \right] dx = e^{ax} f(x) + c$$ | High-frequency FBISE/Punjab Board long & short question pattern (Ex 3.4). |
| Square-Root Quadratic Form | $$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + c$$ | Area under circles, ellipses, and standard board proofs. |
Below are 4 fundamental long questions selected directly from past papers of BISE Lahore, BISE Rawalpindi, and FBISE, formatted step-by-step according to board marking schemes.
Solution:
Let the given integral be:
$$I = \int \sqrt{a^2 - x^2} \cdot 1 \, dx$$Using Integration by Parts $\int u \, dv = u v - \int v \, du$, let:
$$u = \sqrt{a^2 - x^2} \quad \text{and} \quad dv = 1 \cdot dx$$Differentiating $u$ with respect to $x$:
$$\frac{du}{dx} = \frac{1}{2\sqrt{a^2 - x^2}} \cdot \frac{d}{dx}(a^2 - x^2) = \frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}}$$ $$du = \frac{-x}{\sqrt{a^2 - x^2}} \, dx$$Integrating $dv$:
$$v = \int 1 \, dx = x$$Applying the Integration by Parts formula:
$$I = x \sqrt{a^2 - x^2} - \int x \left( \frac{-x}{\sqrt{a^2 - x^2}} \right) dx$$ $$I = x \sqrt{a^2 - x^2} - \int \frac{-x^2}{\sqrt{a^2 - x^2}} \, dx$$Manipulating the numerator by adding and subtracting $a^2$:
$$-x^2 = (a^2 - x^2) - a^2$$ $$I = x \sqrt{a^2 - x^2} - \int \frac{(a^2 - x^2) - a^2}{\sqrt{a^2 - x^2}} \, dx$$ $$I = x \sqrt{a^2 - x^2} - \left( \int \frac{a^2 - x^2}{\sqrt{a^2 - x^2}} \, dx - \int \frac{a^2}{\sqrt{a^2 - x^2}} \, dx \right)$$ $$I = x \sqrt{a^2 - x^2} - \int \sqrt{a^2 - x^2} \, dx + a^2 \int \frac{1}{\sqrt{a^2 - x^2}} \, dx$$Recognizing that $\int \sqrt{a^2 - x^2} \, dx = I$ and $\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right)$:
$$I = x \sqrt{a^2 - x^2} - I + a^2 \sin^{-1}\left(\frac{x}{a}\right)$$ $$I + I = x \sqrt{a^2 - x^2} + a^2 \sin^{-1}\left(\frac{x}{a}\right)$$ $$2I = x \sqrt{a^2 - x^2} + a^2 \sin^{-1}\left(\frac{x}{a}\right)$$ $$I = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + c$$Hence Proved.
Solution:
Let the given integral be:
$$I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx \quad \text{--- (Equation 1)}$$Using the Definite Integral property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$, substitute $x$ with $\left(\frac{\pi}{2} - x\right)$:
$$I = \int_{0}^{\pi/2} \frac{\sin\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)} \, dx$$Using trigonometric co-function identities $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\frac{\pi}{2} - x\right) = \sin x$:
$$I = \int_{0}^{\pi/2} \frac{\cos x}{\cos x + \sin x} \, dx \quad \text{--- (Equation 2)}$$Adding Equation 1 and Equation 2:
$$I + I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx + \int_{0}^{\pi/2} \frac{\cos x}{\sin x + \cos x} \, dx$$ $$2I = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx$$ $$2I = \int_{0}^{\pi/2} 1 \, dx$$ $$2I = [x]_{0}^{\pi/2}$$ $$2I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$ $$I = \frac{\pi}{4}$$Final Answer: $\frac{\pi}{4}$
Solution:
Let $I = \int e^x \left(\frac{1+x}{(2+x)^2}\right) dx$.
Manipulate the algebraic numerator in terms of the denominator $(2+x)$:
$$1+x = (2+x) - 1$$ $$I = \int e^x \left[ \frac{(2+x) - 1}{(2+x)^2} \right] dx$$ $$I = \int e^x \left[ \frac{2+x}{(2+x)^2} - \frac{1}{(2+x)^2} \right] dx$$ $$I = \int e^x \left[ \frac{1}{2+x} + \left( -\frac{1}{(2+x)^2} \right) \right] dx$$Define the function $f(x)$:
$$f(x) = \frac{1}{2+x} = (2+x)^{-1}$$Differentiating $f(x)$ with respect to $x$ gives:
$$f'(x) = -1(2+x)^{-2} \cdot \frac{d}{dx}(2+x) = -\frac{1}{(2+x)^2}$$The integral now perfectly matches the standard form theorem $\int e^{ax} \left[ a f(x) + f'(x) \right] dx = e^{ax} f(x) + c$, where $a = 1$.
Therefore:
$$I = e^{1 \cdot x} f(x) + c$$ $$I = e^x \cdot \left(\frac{1}{2+x}\right) + c$$ $$I = \frac{e^x}{2+x} + c$$Final Answer: $\frac{e^x}{2+x} + c$
Solution:
To determine the limits of integration, find the x-intercepts of the curve by setting $y = 0$:
$$4 - x^2 = 0 \implies x^2 = 4 \implies x = \pm 2$$The lower limit is $a = -2$ and the upper limit is $b = 2$. Throughout the interval $[-2, 2]$, $y \ge 0$.
The total area $A$ bounded by the curve and the x-axis is given by:
$$A = \int_{-2}^{2} y \, dx = \int_{-2}^{2} (4 - x^2) \, dx$$Since $f(x) = 4 - x^2$ is an even function ($f(-x) = 4 - (-x)^2 = 4 - x^2 = f(x)$), we can apply symmetric limits property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$:
$$A = 2 \int_{0}^{2} (4 - x^2) \, dx$$Integrating term by term:
$$A = 2 \left[ 4x - \frac{x^3}{3} \right]_{0}^{2}$$Applying the upper and lower limits:
$$A = 2 \left[ \left( 4(2) - \frac{2^3}{3} \right) - \left( 4(0) - \frac{0^3}{3} \right) \right]$$ $$A = 2 \left[ 8 - \frac{8}{3} \right] = 2 \left[ \frac{24 - 8}{3} \right]$$ $$A = 2 \left[ \frac{16}{3} \right] = \frac{32}{3} \text{ square units}$$Final Answer: $\frac{32}{3}$ square units
Test your conceptual speed for board MCQs and ECAT/NET entry test papers:
Rationale: This is the fundamental Logarithmic Integration Rule. When the numerator is the exact derivative of the denominator, the integral yields the natural logarithm of the absolute value of the denominator.
Rationale: Multiplying and dividing by $(\sec x + \tan x)$ yields $\int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, dx$. Since the numerator is the derivative of the denominator, it integrates to $\ln|\sec x + \tan x| + c$.
Rationale: The antiderivative of $\frac{1}{1+x^2}$ is $\tan^{-1}(x)$. Evaluating from $0$ to $1$: $\tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.
Rationale: Compare with $\int e^{ax}[a f(x) + f'(x)] dx = e^{ax} f(x) + c$. Here $a=2$, $f(x) = \cos x$, and $f'(x) = -\sin x$. Thus, the answer is $e^{2x}\cos x + c$.
Rationale: Let $g(x) = x^3 \cos x$. Then $g(-x) = (-x)^3 \cos(-x) = -x^3 \cos x = -g(x)$. Since $g(x)$ is an odd function, its integral over symmetric limits $[-a, a]$ is identically $0$.
Rationale: Rewrite the integrand as $\frac{1/x}{\ln x}$. The numerator $\frac{1}{x}$ is the derivative of the denominator $\ln x$. Applying $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + c$ yields $\ln|\ln x| + c$.
Rationale: Since $\frac{d}{dx}(a^x) = a^x \ln a$, dividing both sides by the constant $\ln a$ gives the integral $\int a^x dx = \frac{a^x}{\ln a} + c$.
Rationale: Separating variables gives $\frac{1}{y} dy = 1 dx$. Integrating both sides yields $\ln|y| = x + k \implies y = e^{x+k} = e^k \cdot e^x = c e^x$.
Answer: In Pakistani Board syllabi and standard calculus, follow the ILATE rule priority order to select the first function $u$:
The function appearing earlier in ILATE is chosen as $u$ (to be differentiated), while the latter function is chosen as $dv/dx$ (to be integrated).
Answer: When evaluating a definite integral $\int_{a}^{b} f(x) \, dx = [F(x) + c]_{a}^{b}$, we substitute upper and lower limits:
$$\left[ F(b) + c \right] - \left[ F(a) + c \right] = F(b) + c - F(a) - c = F(b) - F(a)$$The arbitrary constant $c$ cancels out subtractively, rendering it redundant for definite boundary evaluations.
Answer: Students frequently apply $\int [f(x)]^n \, dx = \frac{[f(x)]^{n+1}}{n+1}$ without verifying whether the exact differential derivative $f'(x)$ is present alongside the function. The rule is ONLY valid when the integrand contains the exact derivative $f'(x)$ as a multiplicative factor: $\int [f(x)]^n f'(x) \, dx$. If $f'(x)$ is a constant missing factor, it must be multiplied and divided outside the integral before applying the power rule.