Apply integration techniques, including substitution, integration by parts, partial fractions, and properties of definite integrals, to solve mathematical problems.
Integral calculus studies accumulation and the recovery of a function from its rate of change. Its governing principle is the Fundamental Theorem of Calculus: differentiation and integration are inverse processes under suitable continuity conditions.
A function \( F \) is an antiderivative of \( f \) on an interval if \( F'(x) = f(x) \).
The expression \( \int f(x)\,dx \) represents the family of all antiderivatives of \( f \), not a single numerical value.
If \( f \) is continuous on \( [a, b] \), then:
General rules of integration reverse familiar differentiation formulas and allow complicated integrands to be separated into simpler terms.
| Rule | Formula |
|---|---|
| Constant Rule | \( \displaystyle\int k\,dx = kx + C \) |
| Power Rule | \( \displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \) for \( n \neq -1 \) |
| Logarithmic Rule | \( \displaystyle\int \frac{1}{x}\,dx = \ln|x| + C \) for \( x \neq 0 \) |
| Constant Multiple Rule | \( \displaystyle\int kf(x)\,dx = k\int f(x)\,dx \) |
| Sum / Difference Rule | \( \displaystyle\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx \) |
Problem: Integrate \( 3x^2 - 4x + 5 \)
Solution:
\[ \int (3x^2 - 4x + 5)\,dx = 3\cdot\frac{x^3}{3} - 4\cdot\frac{x^2}{2} + 5x + C \] \[ = x^3 - 2x^2 + 5x + C \]Verification: Differentiating the result gives \( 3x^2 - 4x + 5 \) ✓
Problem: Evaluate \( \int (2\sin x + 3\cos x)\,dx \)
Solution:
\[ \int (2\sin x + 3\cos x)\,dx = 2(-\cos x) + 3(\sin x) + C = -2\cos x + 3\sin x + C \]Problem: Evaluate \( \int (e^x + 2^x)\,dx \)
Solution:
\[ \int (e^x + 2^x)\,dx = e^x + \frac{2^x}{\ln 2} + C \]Integration by substitution (also called u-substitution) is the reverse of the chain rule. It transforms a complicated integral into a simpler one by changing the variable of integration.
Problem: Evaluate \( \displaystyle\int (2x + 1)^5\,dx \)
Solution:
Let \( u = 2x + 1 \), then \( du = 2\,dx \) → \( dx = \frac{du}{2} \)
\[ \int (2x+1)^5\,dx = \int u^5 \cdot \frac{du}{2} = \frac{1}{2}\cdot\frac{u^6}{6} + C = \frac{(2x+1)^6}{12} + C \]Problem: Evaluate \( \displaystyle\int \sin(3x)\,dx \)
Solution:
Let \( u = 3x \), then \( du = 3\,dx \) → \( dx = \frac{du}{3} \)
\[ \int \sin(3x)\,dx = \int \sin u \cdot \frac{du}{3} = -\frac{\cos u}{3} + C = -\frac{\cos(3x)}{3} + C \]Problem: Evaluate \( \displaystyle\int \frac{2x}{x^2 + 1}\,dx \)
Solution:
Let \( u = x^2 + 1 \), then \( du = 2x\,dx \)
\[ \int \frac{2x}{x^2 + 1}\,dx = \int \frac{du}{u} = \ln|u| + C = \ln|x^2 + 1| + C \]Problem: Evaluate \( \displaystyle\int_0^1 x(x^2 + 1)^3\,dx \)
Solution:
Let \( u = x^2 + 1 \), then \( du = 2x\,dx \) → \( x\,dx = \frac{du}{2} \)
When \( x = 0 \), \( u = 1 \); when \( x = 1 \), \( u = 2 \)
\[ \int_0^1 x(x^2+1)^3\,dx = \int_1^2 u^3 \cdot \frac{du}{2} = \frac{1}{2}\left[\frac{u^4}{4}\right]_1^2 = \frac{1}{8}(16 - 1) = \frac{15}{8} \]Integration by parts is the reverse of the product rule for differentiation. It is used when the integrand is a product of two functions that cannot be integrated by simple substitution.
where \( u \) and \( v \) are differentiable functions of \( x \).
Use the ILATE mnemonic to choose \( u \) (the function to differentiate):
| Priority | Function Type | Examples |
|---|---|---|
| 1 | I – Inverse trigonometric | \( \sin^{-1}x, \tan^{-1}x \) |
| 2 | L – Logarithmic | \( \ln x, \log_a x \) |
| 3 | A – Algebraic | \( x, x^2, x^3 \) |
| 4 | T – Trigonometric | \( \sin x, \cos x \) |
| 5 | E – Exponential | \( e^x, a^x \) |
Choose \( u \) as the function that appears higher in the list. The remaining function becomes \( dv \).
Problem: Evaluate \( \displaystyle\int x e^x\,dx \)
Solution:
Choose \( u = x \) (Algebraic), \( dv = e^x\,dx \) (Exponential)
Then \( du = dx \), \( v = e^x \)
\[ \int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x - 1) + C \]Problem: Evaluate \( \displaystyle\int x \sin x\,dx \)
Solution:
Choose \( u = x \) (Algebraic), \( dv = \sin x\,dx \) (Trigonometric)
Then \( du = dx \), \( v = -\cos x \)
\[ \int x \sin x\,dx = x(-\cos x) - \int (-\cos x)\,dx = -x\cos x + \sin x + C \]Problem: Evaluate \( \displaystyle\int \ln x\,dx \)
Solution:
Choose \( u = \ln x \) (Logarithmic), \( dv = dx \) (Algebraic)
Then \( du = \frac{1}{x}\,dx \), \( v = x \)
\[ \int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C \]Problem: Evaluate \( \displaystyle\int x^2 e^x\,dx \)
Solution:
First application: \( u = x^2 \), \( dv = e^x\,dx \) → \( du = 2x\,dx \), \( v = e^x \)
\[ \int x^2 e^x\,dx = x^2 e^x - 2\int x e^x\,dx \]Second application (from Example 1): \( \int x e^x\,dx = e^x(x-1) \)
\[ = x^2 e^x - 2e^x(x-1) + C = e^x(x^2 - 2x + 2) + C \]Problem: Evaluate \( \displaystyle\int \tan^{-1}x\,dx \)
Solution:
Choose \( u = \tan^{-1}x \), \( dv = dx \) → \( du = \frac{1}{1+x^2}\,dx \), \( v = x \)
\[ \int \tan^{-1}x\,dx = x\tan^{-1}x - \int \frac{x}{1+x^2}\,dx \]For the remaining integral, use substitution \( t = 1+x^2 \), \( dt = 2x\,dx \):
\[ = x\tan^{-1}x - \frac{1}{2}\ln(1+x^2) + C \]Integration by partial fractions is used to integrate rational functions of the form \( \dfrac{P(x)}{Q(x)} \), where the degree of \( P(x) \) is less than the degree of \( Q(x) \).
The rational function is decomposed into a sum of simpler fractions that can be integrated using standard rules.
| Denominator Factor | Partial Fraction Form |
|---|---|
| Linear: \( (ax + b) \) | \( \dfrac{A}{ax + b} \) |
| Repeated Linear: \( (ax + b)^n \) | \( \dfrac{A_1}{ax+b} + \dfrac{A_2}{(ax+b)^2} + \cdots + \dfrac{A_n}{(ax+b)^n} \) |
| Irreducible Quadratic: \( ax^2 + bx + c \) | \( \dfrac{Ax + B}{ax^2 + bx + c} \) |
| Repeated Quadratic | \( \dfrac{A_1x + B_1}{ax^2+bx+c} + \dfrac{A_2x + B_2}{(ax^2+bx+c)^2} + \cdots \) |
Problem: Evaluate \( \displaystyle\int \frac{1}{(x+1)(x+2)}\,dx \)
Solution:
Step 1: Partial fraction decomposition:
\[ \frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \]Step 2: \( 1 = A(x+2) + B(x+1) \)
Put \( x = -1 \): \( 1 = A(1) \Rightarrow A = 1 \)
Put \( x = -2 \): \( 1 = B(-1) \Rightarrow B = -1 \)
Step 3:
\[ \int \frac{1}{(x+1)(x+2)}\,dx = \int \left(\frac{1}{x+1} - \frac{1}{x+2}\right)dx = \ln|x+1| - \ln|x+2| + C \] \[ = \ln\left|\frac{x+1}{x+2}\right| + C \]Problem: Evaluate \( \displaystyle\int \frac{2x+3}{(x-1)^2}\,dx \)
Solution:
Step 1: Decomposition:
\[ \frac{2x+3}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2} \]Step 2: \( 2x+3 = A(x-1) + B \)
Put \( x = 1 \): \( 5 = B \Rightarrow B = 5 \)
Compare coefficients of \( x \): \( 2 = A \Rightarrow A = 2 \)
Step 3:
\[ \int \frac{2x+3}{(x-1)^2}\,dx = \int \left(\frac{2}{x-1} + \frac{5}{(x-1)^2}\right)dx \] \[ = 2\ln|x-1| - \frac{5}{x-1} + C \]Problem: Evaluate \( \displaystyle\int \frac{x+1}{x^2+1}\,dx \)
Solution:
Step 1: Decomposition:
\[ \frac{x+1}{x^2+1} = \frac{Ax+B}{x^2+1} \]So \( x+1 = Ax + B \Rightarrow A = 1, B = 1 \)
Step 2:
\[ \int \frac{x+1}{x^2+1}\,dx = \int \frac{x}{x^2+1}\,dx + \int \frac{1}{x^2+1}\,dx \] \[ = \frac{1}{2}\ln(x^2+1) + \tan^{-1}x + C \]| Property | Formula |
|---|---|
| Zero Width | \( \displaystyle\int_a^a f(x)\,dx = 0 \) |
| Reversal of Limits | \( \displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx \) |
| Constant Multiple | \( \displaystyle\int_a^b kf(x)\,dx = k\int_a^b f(x)\,dx \) |
| Sum / Difference | \( \displaystyle\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx \) |
| Additivity | \( \displaystyle\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx \) |
| Dummy Variable | \( \displaystyle\int_a^b f(x)\,dx = \int_a^b f(t)\,dt \) |
| Property | Formula |
|---|---|
| King Property | \( \displaystyle\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \) |
| Periodic Property | \( \displaystyle\int_0^{nT} f(x)\,dx = n\int_0^T f(x)\,dx \) for periodic \( f \) with period \( T \) |
| Comparison | If \( f(x) \leq g(x) \) on \( [a,b] \), then \( \displaystyle\int_a^b f(x)\,dx \leq \int_a^b g(x)\,dx \) |
Problem: Evaluate \( \displaystyle\int_{-2}^{2} x^2\,dx \)
Solution:
\( f(x) = x^2 \) is even, so:
\[ \int_{-2}^{2} x^2\,dx = 2\int_0^2 x^2\,dx = 2\left[\frac{x^3}{3}\right]_0^2 = 2\cdot\frac{8}{3} = \frac{16}{3} \]Problem: Evaluate \( \displaystyle\int_{-1}^{1} x^3\,dx \)
Solution:
\( f(x) = x^3 \) is odd, so:
\[ \int_{-1}^{1} x^3\,dx = 0 \]Problem: Evaluate \( \displaystyle\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx \)
Solution:
Let \( I = \displaystyle\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx \)
By the king property:
\[ I = \int_0^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)}\,dx = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x}\,dx \]Adding the two expressions for \( I \):
\[ 2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2} \] \[ \boxed{I = \frac{\pi}{4}} \]Problem: Evaluate \( \displaystyle\int_0^1 \frac{1}{1+x^2}\,dx \)
Solution:
\[ \int_0^1 \frac{1}{1+x^2}\,dx = [\tan^{-1}x]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]| Topic | Key Formula / Concept |
|---|---|
| Indefinite Integral | \( \int f(x)\,dx = F(x) + C \) |
| Definite Integral | \( \int_a^b f(x)\,dx = F(b) - F(a) \) |
| Power Rule | \( \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \) (\( n \neq -1 \)) |
| Logarithmic Rule | \( \int \frac{1}{x}\,dx = \ln|x| + C \) |
| Exponential Rule | \( \int e^x\,dx = e^x + C \), \( \int a^x\,dx = \frac{a^x}{\ln a} + C \) |
| Trigonometric Rules | \( \int \sin x\,dx = -\cos x + C \), \( \int \cos x\,dx = \sin x + C \) |
| Substitution | \( \int f(g(x))g'(x)\,dx = \int f(u)\,du \) |
| Integration by Parts | \( \int u\,dv = uv - \int v\,du \) |
| ILATE Rule | Priority: Inverse → Log → Algebraic → Trig → Exponential |
| Partial Fractions | \( \frac{P(x)}{Q(x)} = \sum \frac{A_i}{(a_i x + b_i)} + \sum \frac{B_j x + C_j}{(\text{quadratic})_j} \) |
| Zero Width | \( \int_a^a f(x)\,dx = 0 \) |
| Reversal | \( \int_a^b f(x)\,dx = -\int_b^a f(x)\,dx \) |
| Additivity | \( \int_a^b f = \int_a^c f + \int_c^b f \) |
| Even Function | \( \int_{-a}^{a} f = 2\int_0^a f \) |
| Odd Function | \( \int_{-a}^{a} f = 0 \) |
| King Property | \( \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \) |
| Integral | Result |
|---|---|
| \( \int x^n\,dx \) | \( \frac{x^{n+1}}{n+1} + C \) (\( n \neq -1 \)) |
| \( \int \frac{1}{x}\,dx \) | \( \ln|x| + C \) |
| \( \int e^x\,dx \) | \( e^x + C \) |
| \( \int a^x\,dx \) | \( \frac{a^x}{\ln a} + C \) |
| \( \int \sin x\,dx \) | \( -\cos x + C \) |
| \( \int \cos x\,dx \) | \( \sin x + C \) |
| \( \int \sec^2 x\,dx \) | \( \tan x + C \) |
| \( \int \csc^2 x\,dx \) | \( -\cot x + C \) |
| \( \int \sec x \tan x\,dx \) | \( \sec x + C \) |
| \( \int \csc x \cot x\,dx \) | \( -\csc x + C \) |
| \( \int \frac{1}{1+x^2}\,dx \) | \( \tan^{-1}x + C \) |
| \( \int \frac{1}{\sqrt{1-x^2}}\,dx \) | \( \sin^{-1}x + C \) |
| \( \int \frac{1}{x^2+a^2}\,dx \) | \( \frac{1}{a}\tan^{-1}\frac{x}{a} + C \) |
| \( \int \frac{1}{\sqrt{a^2-x^2}}\,dx \) | \( \sin^{-1}\frac{x}{a} + C \) |
Evaluate \( \displaystyle\int (3x^4 - 2x^2 + 7)\,dx \)
Evaluate \( \displaystyle\int x(x^2 + 3)^4\,dx \)
Evaluate \( \displaystyle\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx \)
Evaluate \( \displaystyle\int x \cos x\,dx \)
Evaluate \( \displaystyle\int x^2 \ln x\,dx \)
Evaluate \( \displaystyle\int \frac{1}{x^2 - 1}\,dx \)
Evaluate \( \displaystyle\int_{-1}^{1} (x^3 + 2x)\,dx \)
Evaluate \( \displaystyle\int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x}\,dx \)
Let \( u = x^2 + 3 \), then \( du = 2x\,dx \) → \( x\,dx = \frac{du}{2} \)
\[ \int x(x^2+3)^4\,dx = \int u^4 \cdot \frac{du}{2} = \frac{1}{2}\cdot\frac{u^5}{5} + C = \frac{(x^2+3)^5}{10} + C \]Let \( u = \sqrt{x} \), then \( du = \frac{1}{2\sqrt{x}}\,dx \) → \( \frac{dx}{\sqrt{x}} = 2\,du \)
\[ \int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx = \int e^u \cdot 2\,du = 2e^u + C = 2e^{\sqrt{x}} + C \]\( u = x \), \( dv = \cos x\,dx \) → \( du = dx \), \( v = \sin x \)
\[ \int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C \]\( u = \ln x \), \( dv = x^2\,dx \) → \( du = \frac{1}{x}\,dx \), \( v = \frac{x^3}{3} \)
\[ \int x^2\ln x\,dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3}\cdot\frac{1}{x}\,dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx \] \[ = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C = \frac{x^3}{3}\left(\ln x - \frac{1}{3}\right) + C \]\( \frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \)
\( 1 = A(x+1) + B(x-1) \)
\( x = 1 \): \( 1 = 2A \Rightarrow A = \frac{1}{2} \)
\( x = -1 \): \( 1 = -2B \Rightarrow B = -\frac{1}{2} \)
\[ \int \frac{1}{x^2-1}\,dx = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C \]\( \int_{-1}^{1} (x^3 + 2x)\,dx \)
\( x^3 \) is odd, \( 2x \) is odd → \( x^3 + 2x \) is odd
For an odd function, \( \int_{-a}^{a} f(x)\,dx = 0 \)
\[ \boxed{0} \]\( I = \displaystyle\int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x}\,dx \)
By King Property:
\[ I = \int_0^{\pi/2} \frac{\cos(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)}\,dx = \int_0^{\pi/2} \frac{\sin x}{\cos x + \sin x}\,dx \]Adding:
\[ 2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2} \] \[ \boxed{I = \frac{\pi}{4}} \]| Reference | Book |
|---|---|
| T-1 | Advanced Engineering Mathematics – R.K. Jain, S.R.K. Iyengar (Narosa Publishing House) |
| R-1 | Higher Engineering Mathematics – B.S. Grewal (Khanna Publishers) |
| R-2 | Mathematics Textbook for Class XII Part I – NCERT |
| R-3 | Mathematics Textbook for Class XII Part II – NCERT |
| OR-1 | https://ncert.nic.in/textbook.php?lemh1=5-6 |
| OR-2 | https://ncert.nic.in/textbook.php?lemh2=1-7 |
After studying this unit, students should be able to:
| CO3 Components | Covered in Section |
|---|---|
| Apply substitution techniques | III |
| Apply integration by parts | IV |
| Apply partial fractions | V |
| Use properties of definite integrals | VI |
| Evaluate standard integrals | II |
⭐ End of Unit III Notes ⭐
Fundamentals of Integral Calculus
MTH165 – Mathematics for Engineers
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