Class 12 FSc Math Chapter 3 Integration Formulas & Solved Board Questions PDF | BISE Punjab & FBISE Study Guide
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).
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.
The Fundamental Theorem of Calculus bridges differential calculus and integral calculus, consisting of two main parts:
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.
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 |
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: $\int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} [a \sin(bx) - b \cos(bx)] + C$
---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: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$
---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.
---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 $y - x = C_1 (1 + xy)$.
Q1: What is the integral $\int \frac{1}{x \ln(x)} \, dx$?
Q2: Evaluate the integral $\int e^x \left( \tan^{-1} x + \frac{1}{1+x^2} \right) dx$.
Q3: What is the value of the definite integral $\int_{-\pi}^{\pi} \sin^3(x) \cos^2(x) \, dx$?
Q4: Evaluate $\int \frac{1}{\sqrt{16 - x^2}} \, dx$.
Q5: What is the order and degree of the differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + 4y = 0$?
Q6: Evaluate $\int_0^2 (3x^2 - 2x + 1) \, dx$.
Q7: What is $\int \sec(x) \, dx$?
Q8: If $f'(x) = 4x^3 - 3x^2$ and $f(1) = 0$, find $f(x)$.
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.
To choose $u(x)$ effectively, follow the LIATE priority rule (from top priority for $u$ to lowest):
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).
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.