We will explore fundamental integration rules used in calculus, and cover essential concepts such as the constant rule, exponential rule, power rule, and trigonometric rules. Each rule is illustrated with practical examples, helping readers gain confidence in integrating various functions. This document serves as a valuable reference for understanding and applying integration techniques.
Trig substitution is particularly useful when dealing with integrals containing the following types of expressions:
The choice of trigonometric substitution depends on the form of the integral. Here are the common substitutions:
Form Substitution a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2 x = a sin θ x = a\sin\theta x = a sin θ x 2 − a 2 \sqrt{x^2 - a^2} x 2 − a 2 x = a cos θ x = a\cos\theta x = a cos θ x 2 + a 2 \sqrt{x^2 + a^2} x 2 + a 2 x = a tan θ x = a\tan\theta x = a tan θ a 2 − x 2 a^2 - x^2 a 2 − x 2 x = a sin θ x = a\sin\theta x = a sin θ a 2 + x 2 a^2 + x^2 a 2 + x 2 x = a tan θ x = a\tan\theta x = a tan θ x 2 − a 2 x^2 - a^2 x 2 − a 2 x = a sec θ x = a\sec\theta x = a sec θ Rational expressions with quadratic denominators Use appropriate substitution based on the form
Evaluate ∫ 1 4 − x 2 d x \displaystyle\int \frac{1}{\sqrt{4 - x^2}}\,dx ∫ 4 − x 2 1 d x
Solution: Let x = 2 sin θ x = 2\sin\theta x = 2 sin θ d x = 2 cos θ d θ dx = 2\cos\theta\,d\theta d x = 2 cos θ d θ
∫ 1 4 − x 2 d x = ∫ 1 4 − ( 2 sin θ ) 2 ⋅ 2 cos θ d θ = ∫ 1 4 − 4 sin 2 θ ⋅ 2 cos θ d θ \int \frac{1}{\sqrt{4 - x^2}}\,dx = \int \frac{1}{\sqrt{4 - (2\sin\theta)^2}} \cdot 2\cos\theta\,d\theta = \int \frac{1}{\sqrt{4 - 4\sin^2\theta}} \cdot 2\cos\theta\,d\theta ∫ 4 − x 2 1 d x = ∫ 4 − ( 2 sin θ ) 2 1 ⋅ 2 cos θ d θ = ∫ 4 − 4 sin 2 θ 1 ⋅ 2 cos θ d θ = ∫ 1 4 cos 2 θ ⋅ 2 cos θ d θ = ∫ 1 2 d θ = θ 2 + C = arcsin ( x 2 ) 2 + C = \int \frac{1}{\sqrt{4\cos^2\theta}} \cdot 2\cos\theta\,d\theta = \int \frac{1}{2}\,d\theta = \frac{\theta}{2} + C = \frac{\arcsin\!\left(\frac{x}{2}\right)}{2} + C = ∫ 4 cos 2 θ 1 ⋅ 2 cos θ d θ = ∫ 2 1 d θ = 2 θ + C = 2 arcsin ( 2 x ) + C Evaluate ∫ 1 x 2 − 1 d x \displaystyle\int \frac{1}{\sqrt{x^2 - 1}}\,dx ∫ x 2 − 1 1 d x
Solution: Let x = sec θ x = \sec\theta x = sec θ d x = sec θ tan θ d θ dx = \sec\theta\tan\theta\,d\theta d x = sec θ tan θ d θ
∫ 1 x 2 − 1 d x = ∫ 1 sec 2 θ − 1 ⋅ sec θ tan θ d θ = ∫ 1 tan 2 θ ⋅ sec θ tan θ d θ \int \frac{1}{\sqrt{x^2 - 1}}\,dx = \int \frac{1}{\sqrt{\sec^2\theta - 1}} \cdot \sec\theta\tan\theta\,d\theta = \int \frac{1}{\sqrt{\tan^2\theta}} \cdot \sec\theta\tan\theta\,d\theta ∫ x 2 − 1 1 d x = ∫ sec 2 θ − 1 1 ⋅ sec θ tan θ d θ = ∫ tan 2 θ 1 ⋅ sec θ tan θ d θ = ∫ 1 ∣ tan θ ∣ ⋅ sec θ tan θ d θ = ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C = ln ∣ x + x 2 − 1 ∣ + C = \int \frac{1}{|\tan\theta|} \cdot \sec\theta\tan\theta\,d\theta = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C = \ln\left|x + \sqrt{x^2 - 1}\right| + C = ∫ ∣ tan θ ∣ 1 ⋅ sec θ tan θ d θ = ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C = ln x + x 2 − 1 + C Evaluate ∫ 1 x 2 1 − x 2 d x \displaystyle\int \frac{1}{x^2\sqrt{1 - x^2}}\,dx ∫ x 2 1 − x 2 1 d x
Solution: Let x = sin θ x = \sin\theta x = sin θ d x = cos θ d θ dx = \cos\theta\,d\theta d x = cos θ d θ
∫ 1 x 2 1 − x 2 d x = ∫ 1 ( sin θ ) 2 1 − ( sin θ ) 2 ⋅ cos θ d θ = ∫ 1 sin 2 θ cos θ ⋅ cos θ d θ \int \frac{1}{x^2\sqrt{1 - x^2}}\,dx = \int \frac{1}{(\sin\theta)^2\sqrt{1 - (\sin\theta)^2}} \cdot \cos\theta\,d\theta = \int \frac{1}{\sin^2\theta\cos\theta} \cdot \cos\theta\,d\theta ∫ x 2 1 − x 2 1 d x = ∫ ( sin θ ) 2 1 − ( sin θ ) 2 1 ⋅ cos θ d θ = ∫ sin 2 θ cos θ 1 ⋅ cos θ d θ = ∫ 1 sin 2 θ d θ = ∫ csc 2 θ d θ = − cot θ + C = − cot ( sin − 1 x ) + C = \int \frac{1}{\sin^2\theta}\,d\theta = \int \csc^2\theta\,d\theta = -\cot\theta + C = -\cot(\sin^{-1} x) + C = ∫ sin 2 θ 1 d θ = ∫ csc 2 θ d θ = − cot θ + C = − cot ( sin − 1 x ) + C Evaluate ∫ 1 x 2 + 9 d x \displaystyle\int \frac{1}{\sqrt{x^2 + 9}}\,dx ∫ x 2 + 9 1 d x
Solution: Let x = 3 tan θ x = 3\tan\theta x = 3 tan θ d x = 3 sec 2 θ d θ dx = 3\sec^2\theta\,d\theta d x = 3 sec 2 θ d θ
∫ 1 x 2 + 9 d x = ∫ 1 ( 3 tan θ ) 2 + 9 ⋅ 3 sec 2 θ d θ = ∫ 1 9 tan 2 θ + 9 ⋅ 3 sec 2 θ d θ \int \frac{1}{\sqrt{x^2 + 9}}\,dx = \int \frac{1}{\sqrt{(3\tan\theta)^2 + 9}} \cdot 3\sec^2\theta\,d\theta = \int \frac{1}{\sqrt{9\tan^2\theta + 9}} \cdot 3\sec^2\theta\,d\theta ∫ x 2 + 9 1 d x = ∫ ( 3 tan θ ) 2 + 9 1 ⋅ 3 sec 2 θ d θ = ∫ 9 tan 2 θ + 9 1 ⋅ 3 sec 2 θ d θ = ∫ 1 9 ( tan 2 θ + 1 ) ⋅ 3 sec 2 θ d θ = ∫ 1 9 sec 2 θ ⋅ 3 sec 2 θ d θ = ∫ 1 d θ = θ + C = arctan ( x 3 ) + C = \int \frac{1}{\sqrt{9(\tan^2\theta + 1)}} \cdot 3\sec^2\theta\,d\theta = \int \frac{1}{\sqrt{9\sec^2\theta}} \cdot 3\sec^2\theta\,d\theta = \int 1\,d\theta = \theta + C = \arctan\!\left(\frac{x}{3}\right) + C = ∫ 9 ( tan 2 θ + 1 ) 1 ⋅ 3 sec 2 θ d θ = ∫ 9 sec 2 θ 1 ⋅ 3 sec 2 θ d θ = ∫ 1 d θ = θ + C = arctan ( 3 x ) + C Status Source Problem Name Difficulty Tags
