Title 74 Integration Techniques.pdf ✅

MSTC 74: Integration Techniques 1. Integration by Substitution This method features a change in variables in the integral to transform it into an equivalent but simpler version and to solve the integral using elementary methods. 1.1. Substitution suggested by the integral This technique does not have a generalized formula or rule. The substitution comes from identifying possible functions and then checking their feasibility to change the variables in the integral. - Example: ∫ x√x − 1 dx [SOLUTION] From the integral, a possible substitution is u = √x − 1. Manipulating the substitution, u 2 = x − 1 x = u 2 − 1 dx = 2u du Which is possible since ∫ x√x − 1 dx = ∫(u 2 − 1)(u)(2u)du = ∫(2u 4 − 2u 2 )du = 2 5 u 5 − 2 3 u 3 + C = 2 15 u 3 (3u 2 − 5) + C ∫ x√x − 1 dx = 2 15 (x − 1) 3 2(3x − 8) + C
1.2. Trigonometric/Hyperbolic Substitution The integral does not directly suggest this technique, but specific portions may give hints. - Example:∫√1 − x 2 dx [SOLUTION by Trigonometric Substitution] Recall that sin2 θ + cos2 θ = 1 cos θ = ±√1 − sin2 θ Which is similar to the integral. Thus, the substitution is x = sin θ which results to dx = cos θ dθ Manipulating the integral, ∫ √1 − x 2dx = ∫ √1 − sin2 θ (cos θ)dθ = ∫ cos2 θ dθ = 1 2 ∫(1 + cos 2θ)dθ = 1 2 θ + 1 4 sin 2θ + C = 1 2 θ + 1 2 sin θ cos θ + C From the substitution, θ = arcsin x cos θ = √1 − x 2 Therefore, ∫ √1 − x 2 dx = 1 2 arcsin x + 1 2 x√1 − x 2 + C - Example: ∫ √1 + x 2dx [SOLUTION by Hyperbolic Substitution] Let x = sinh u → dx = cosh u du ∫ √1 + x 2dx = ∫ √1 + sinh2 u (cosh u)du = ∫ cosh2 u du = 1 2 ∫(1 + cosh 2u)du = 1 2 u + 1 4 sinh 2u + C = 1 2 u + 1 2 sinh u cosh u + C
From the substitution, u = arcsinh x → cosh u = √1 + x 2 Therefore, ∫ √1 + x 2dx = 1 2 arcsinh x + 1 2 x√1 + x 2 + C 1.3. Weierstrass Substitution This technique is a miscellaneous substitution technique that features a reverse effect of trigonometric substitution. Suppose trigonometric substitution changes the integral from algebraic form to trigonometric form. In that case, Weierstrass substitution changes an integral in trigonometric form to algebraic form. The usual substitution is u = tan x 2 . - Example: ∫ dx 1+sinx [SOLUTION] Let u = tan x 2 . x = 2 arctan u dx = 2 1 + u 2 du From trigonometric identities, u 2 = tan2 x 2 = 1 − cos x 1 + cos x Solving for cos x, cos x = 1 − u 2 1 + u 2 From the Pythagorean identities, sin x = 2u 1 + u 2 Therefore, ∫ dx 1 + sin x = ∫ 2 1 + u 2 du 1 + 2u 1 + u 2 = ∫ 2 (1 + u) 2 du = − 1 1 + u + C
∫ dx 1 + sin x = − 1 1 + tan x 2 + C 2. Integration by Parts [DERIVATION] Consider the product rule in differentiation, d(uv) = u dv + v du Rewrite the equation u dv = d(uv) − v du Integrating both sides give ∫ u dv = uv − ∫ v du Thus, to evaluate an integral by integration by parts, identify the functions u and dv. Famous mnemonics to determine the functions for u and dv are LIATE or LIPET. LIATE stands for Logarithmic, Inverse Trig, Algebraic, Trigonometric, Exponential LIPET stands for Logarithmic, Inverse Trig, Polynomial, Exponential, Trigonometric For two functions in the integrand, the “left” function (based on the mnemonic) is the function u, while the “right” function is the dv. - Example: ∫ x sin x dx [SOLUTION] From the mnemomics, x is algebraic and sin x is trigonometric. Thus u = x, dv = sin x. Also, du = 1, v = − cos x ∫ x sin x dx = x(− cos x) − ∫(− cos x)(1)dx ∫ x sin x dx = −x cos x + sin x + C A more visual method for integration by parts is the array method. Using the same example, Sign u dv + x sin x - 1 − cos x + 0 − sin x The sign and u columns multiply horizontally while u and dv multiply diagonally. The bottommost row of u and dv additionally multiplies horizontally but with an integral sign. ∫ x sin x dx = +x(− cos x) − (1)(− sin x) + ∫ 0(sin x)dx ∫ x sin x dx = −x cos x + sin x + C