Calculus II: Chapter 7 Integration Techniques Review
Chapter 7 Review: Comprehensive Integration Techniques
This review covers a wide array of integration techniques, including integration by parts, trigonometric integrals, substitution methods, hyperbolic functions, partial fractions, trigonometric substitution, definite integrals, improper integrals, and numerical approximation.
Indefinite Integrals: Integration by Parts and Related Techniques
Problem Statement Unclear; Result Provided as:
5ye - 25e + CIntegration of
x imes ext{trigonometric function}:Integral: `\int x \sin(\pi x) dx
Solution:
-\frac{1}{\pi}x \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x) + C(Note: The provided solution
(x-1) \cos(\pi x) + \frac{1}{\pi} \sin(\pi x) + Cseems to be an error in transcription or calculation, as standard integration by parts\int u dv = uv - \int v duwithu=x, dv=\sin(\pi x) dxleads to the corrected solution above.)
Integral of
\cos^3 x:Integral: `\int \cos^3 x dx
Given Solution:
x \cos^3 x - (1 - \frac{1}{2}x^2) + C(Note: This solution does not align with standard techniques for
\int \cos^3 x dx, which typically involves\sin x - \frac{1}{3} \sin^3 x + C.)
Integral of
\sin(\sqrt{x})(Substitution and Integration by Parts):Integral: `\int \sin(\sqrt{x}) dx
Solution:
2\sqrt{x}\sin(\sqrt{x}) + 2\cos(\sqrt{x}) + C(Method: Let
u = \sqrt{x}, thenx=u^2anddx = 2u du. The integral becomes2\int u \sin u du, which is solved using integration by parts.)
Integral of
t e^{nt}:Integral: `\int t e^{nt} dt
Solution:
\frac{1}{n} t e^{nt} - \frac{1}{n^2} e^{nt} + C(Method: Integration by parts with
u=t, dv=e^{nt} dt.)
Integral of
(x^2+2x) \cos x(Repeated Integration by Parts):Integral: `\int (x^2+2x) \cos x dx
Solution:
\left(x^2+2x\right) \sin x + (2x+2) \cos x - 2 \sin x + C(Method: Requires two applications of integration by parts, differentiating
x^2+2xand integrating\cos x.)
Integral of
t^2 \sin(\beta t)(Repeated Integration by Parts):Integral: `\int t^2 \sin(\beta t) dt
Solution:
-\frac{t^2}{\beta} \cos(\beta t) + \frac{2t}{\beta^2} \sin(\beta t) + \frac{2}{\beta^3} \cos(\beta t) + C
Integral of
x \cosh(ax)(Integration by Parts with Hyperbolic Functions):Integral: `\int x \cosh(ax) dx
Solution:
\frac{x}{a} \sinh(ax) - \frac{1}{a^2} \cosh(ax) + C
Integral of
y \sinh(2y)(Integration by Parts with Hyperbolic Functions):Integral: `\int y \sinh(2y) dy
Solution:
\frac{y}{2} \cosh(2y) - \frac{1}{4} \sinh(2y) + C
Integral of
x e^{2x}(Integration by Parts):Integral: `\int x e^{2x} dx
Solution:
\frac{e^{2x}}{4}(2x-1) + C(or\frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C)
Integral of
\frac{e^{\sqrt{x}}}{\sqrt{x}}(Substitution):Integral: `\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx
Solution:
2e^{\sqrt{x}} + C(Method: Let
u = \sqrt{x}, thendu = \frac{1}{2\sqrt{x}} dx.)
Integral of
x^2 \ln x(Integration by Parts):Integral: `\int x^2 \ln x dx
Solution:
\frac{x^3}{3} \ln|x| - \frac{x^3}{9} + C
Indefinite Integrals: Trigonometric Powers and Products
Integral of
\tan^2 \theta \sec^4 \theta(Trigonometric Power):Integral: `\int \tan^2 \theta \sec^4 \theta d\theta
Solution:
\frac{1}{5}\tan^5 \theta + \frac{1}{3}\tan^3 \theta + C(Method:
\sec^4 \theta = \sec^2 \theta \cdot \sec^2 \theta = (1+\tan^2 \theta) \sec^2 \theta. Letu=\tan \theta.)
Integral of
\sin^3 \theta \cos^{1/2} \theta(Trigonometric Power):Integral: `\int \sin^3 \theta \cos^{1/2} \theta d\theta
Solution:
-\frac{2}{3} \cos^{3/2} \theta + \frac{2}{7} \cos^{7/2} \theta + C(Method:
\sin^3 \theta = \sin^2 \theta \sin \theta = (1-\cos^2 \theta) \sin \theta. Letu=\cos \theta.)
Integral of
\sin^3(2t) \cos^2(2t)(Trigonometric Power):Integral: `\int \sin^3(2t) \cos^2(2t) dt
Solution:
-\frac{1}{6} \cos^3(2t) + \frac{1}{10} \cos^5(2t) + C(Method: Similar to above,
\sin^3(2t) = (1-\cos^2(2t)) \sin(2t). Letu=\cos(2t).)
Integral of
\sin^3 x \cos^{1/2} x(Trigonometric Power):Integral: `\int \sin^3 x \cos^{1/2} x dx
Solution:
-\frac{2}{3} \cos^{3/2} x + \frac{2}{7} \cos^{7/2} x + C(This appears to be a repeat of problem 15 with
\thetareplaced byx.)
Integral of
\cos^3(2t)(Trigonometric Power):Integral: `\int \cos^3(2t) dt
Solution:
\frac{1}{2} \sin(2t) - \frac{1}{6} \sin^3(2t) + C(Method:
\cos^3(2t) = (1-\sin^2(2t)) \cos(2t). Letu=\sin(2t).)
Integral of
\tan^4 x \sec^6 x(Trigonometric Power):Integral: `\int \tan^4 x \sec^6 x dx
Solution:
\frac{1}{9} \tan^9 x + \frac{2}{7} \tan^7 x + \frac{1}{5} \tan^5 x + C(Method:
\sec^6 x = \sec^4 x \sec^2 x = (1+\tan^2 x)^2 \sec^2 x. Letu=\tan x.)
Indefinite Integrals: Trigonometric Product-to-Sum Identities
Integral of
\sin x \sin(2x):Integral: `\int \sin x \sin(2x) dx
Solution:
\frac{1}{2} \sin^2 x + C(This solution is simplified from using product-to-sum:\frac{1}{2t}\cos t - \frac{1}{6}\cos 3t + Cfor\sin x \sin(2x) = \frac{1}{2}(\cos(x-2x) - \cos(x+2x)) = \frac{1}{2}(\cos(-x) - \cos(3x)) = \frac{1}{2}(\cos x - \cos(3x)))(Correction:
\int \sin x \sin(2x) dx = \int \sin x (2\sin x \cos x) dx = 2 \int \sin^2 x \cos x dx. Letu=\sin x,du=\cos x dx. Then2 \int u^2 du = \frac{2}{3} u^3 + C = \frac{2}{3} \sin^3 x + C. The provided\frac{1}{2}\sin^2 x + Cis likely incorrect for\sin x \sin(2x).)
Integral of
\cos(8x) \cos(5x):Integral: `\int \cos(8x) \cos(5x) dx
Solution:
\frac{1}{26} \sin(13x) + \frac{1}{6} \sin(3x) + C(Method: Use product-to-sum identity:
\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)].)
Integral of
\sin(2\theta) \sin(6\theta):Integral: `\int \sin(2\theta) \sin(6\theta) d\theta
Solution:
-\frac{1}{8} \sin(4\theta) + \frac{1}{16} \sin(8\theta) + C(Method: Use product-to-sum identity:
\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)].)
Indefinite Integrals: Trigonometric Substitution
Integral of
\frac{\sqrt{9-x^2}}{x}:Integral: `\int \frac{\sqrt{9-x^2}}{x} dx
Solution:
\sqrt{9-x^2} - 3 \ln\left|\frac{3+\sqrt{9-x^2}}{x}\right| + C(or\sqrt{9-x^2} - 3 \operatorname{arcsech}(x/3) + C)(Method: Let
x = 3 \sin \theta.)
Integral of
\frac{\sqrt{9-x^2}}{x^2}:Integral: `\int \frac{\sqrt{9-x^2}}{x^2} dx
Solution:
-\frac{\sqrt{9-x^2}}{x} - \arcsin\left(\frac{x}{3}\right) + C(Method: Let
x = 3 \sin \theta.)
Integral of
\frac{dx}{x^2 \sqrt{4-x^2}}:Integral: `\int \frac{dx}{x^2 \sqrt{4-x^2}}
Solution:
-\frac{\sqrt{4-x^2}}{4x} + C(Method: Let
x = 2 \sin \theta.)
Integral of
\frac{dx}{(x^2+4)^{3/2}}:Integral: `\int \frac{dx}{(x^2+4)^{3/2}}
Solution:
\frac{x}{4\sqrt{x^2+4}} + C(Method: Let
x = 2 \tan \theta.)
Integral of
\frac{x^2}{\sqrt{9-x^2}} dx:Integral: `\int \frac{x^2}{\sqrt{9-x^2}} dx
Solution:
\frac{9}{2} \arcsin\left(\frac{x}{3}\right) - \frac{x}{2} \sqrt{9-x^2} + C(Method: Let
x = 3 \sin \theta.)
Integral of
\frac{\sqrt{x^2-1}}{x^4} dx:Integral: `\int \frac{\sqrt{x^2-1}}{x^4} dx
Solution:
\frac{1}{3} \frac{(x^2-1)^{3/2}}{x^3} + C(Method: Let
x = \sec \theta.)
Integral of
\frac{\sqrt{x^2-9}}{x^3} dx:Integral: `\int \frac{\sqrt{x^2-9}}{x^3} dx
Solution:
\frac{1}{6} \operatorname{arcsec}\left(\frac{|x|}{3}\right) - \frac{\sqrt{x^2-9}}{2x^2} + C(Method: Let
x = 3 \sec \theta.)
Indefinite Integrals: Partial Fractions and Completing the Square
Integral of
\frac{x+2}{x^2+2x+5} dx(Completing the Square):Integral: `\int \frac{x+2}{x^2+2x+5} dx
Solution:
\frac{1}{2} \ln|x^2+2x+5| + \frac{1}{2} \arctan\left(\frac{x+1}{2}\right) + C(Method: Complete the square in the denominator
x^2+2x+5 = (x+1)^2+4. Split the numerator into parts that match the derivative of the denominator and a constant for\arctan.)
Integral of
\frac{x^2+2x-1}{x(2x-1)(x+2)} dx(Partial Fractions):Integral: `\int \frac{x^2+2x-1}{x(2x-1)(x+2)} dx
Solution:
\frac{1}{2} \ln|x| + \frac{1}{10} \ln|2x-1| - \frac{1}{5} \ln|x+2| + C
Integral of
\frac{x^3-2x^2+4x+1}{x^2-2x+4} dx(Polynomial Long Division and Partial Fractions):Integral: `\int \frac{x^3-2x^2+4x+1}{x^2-2x+4} dx
Solution:
\frac{x^2}{2} + \ln|x^2-2x+4| + C(Method: Perform polynomial long division first, then integrate the resulting polynomial and the remainder.)
Integral of
\frac{2x^2-x+4}{x(x^2+4)} dx(Partial Fractions):Integral: `\int \frac{2x^2-x+4}{x(x^2+4)} dx
Solution:
\ln|x| + \frac{1}{2}\ln(x^2+4) - \frac{1}{4} \arctan\left(\frac{x}{2}\right) + C
Integral of
\frac{x^4}{x-1} dx(Polynomial Long Division):Integral: `\int \frac{x^4}{x-1} dx
Solution:
\frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + x + \ln|x-1| + C
Integral of
\frac{x-4}{x^2-3x+2} dx(Partial Fractions):Integral: `\int \frac{x-4}{x^2-3x+2} dx
Solution:
3 \ln|x-1| - 2 \ln|x-2| + C(Method: Factor the denominator
x^2-3x+2 = (x-1)(x-2).)
Integral of
\frac{1}{1-\sin x} dx(Conjugate Multiplication):Integral: `\int \frac{1}{1-\sin x} dx
Solution:
\tan x + \sec x + C(Method: Multiply numerator and denominator by
1+\sin xto get\frac{1+\sin x}{\cos^2 x}and split into\sec^2 x + \tan x \sec x.)
Indefinite Integrals: Mixed Techniques
Integral of
t \tan^2 t dt(Integration by Parts and Identities):Integral: `\int t \tan^2 t dt
Solution:
t \tan t - \frac{1}{2} t^2 - \ln|\cos t| + C(Method: Use
\tan^2 t = \sec^2 t - 1. Then apply integration by parts tot (\sec^2 t - 1).)
Integral of
\frac{dx}{x^3 \sqrt{x^2-1}}(Trigonometric Substitution):Integral: `\int \frac{dx}{x^3 \sqrt{x^2-1}}
Solution:
\frac{\sqrt{x^2-1}}{2x^2} + \frac{1}{2} \operatorname{arcsec}|x| + C
Definite Integrals
Definite Integral
\int_{0}^{\pi} t \cos^2 t dt:Integral: `\int_{0}^{\pi} t \cos^2 t dt
Solution:
\frac{\pi^2}{4}(Method: Use half-angle identity
\cos^2 t = \frac{1+\cos(2t)}{2}, then integrate by parts.)
Definite Integral
\int_{0}^{\pi} y \sin^2 y dy:Integral: `\int_{0}^{\pi} y \sin^2 y dy
Solution:
\frac{\pi^2}{4}(Method: Use half-angle identity
\sin^2 y = \frac{1-\cos(2y)}{2}, then integrate by parts.)
Numerical Integration
Approximation Using Simpson's Rule:
Problem: Approximate
\int_{0}^{2} \frac{1}{1+x^2} dxwithn=10Result:
1.10714
Improper Integrals
Integral of
\int_{1}^{\infty} \frac{1}{x^3} dx:Integral: `\int_{1}^{\infty} \frac{1}{x^3} dx
Result:
= \frac{1}{2}(Converges)
Integral of
\int_{2}^{\infty} \frac{1}{x-2^{3/2}} dx:Integral: `\int_{2}^{\infty} \frac{1}{(x-2)^{3/2}} dx
Result: Diverges (
\infty)
Integral of
\int_{0}^{1} \frac{1}{\sqrt{t}} dt:Integral: `\int_{0}^{1} \frac{1}{\sqrt{t}} dt
Result: Converges (
2)
Integral of
\int_{0}^{\infty} y^3 e^{-3y^2} dy:Integral: `\int_{0}^{\infty} y^3 e^{-3y^2} dy
Result: Converges (
\frac{1}{54})(Method: Requires substitution
u=3y^2and then integration by parts.)
Integral of
\int_{0}^{\pi} \frac{1}{\sin^2 \theta} d\theta:Integral: `\int{0}^{\pi} \frac{1}{\sin^2 \theta} d\theta = \int{0}^{\pi} \csc^2 \theta d\theta
Result: Diverges
(Explanation:
\csc^2 \thetahas vertical asymptotes at\theta=0and\theta=\pi, making it an improper integral.\int \csc^2 \theta d\theta = -\cot \theta. Evaluating-\cot \thetaat0and\pishows divergence.)