# Calculus II For Dummies

By Mark Zegarelli

**An easy-to-understand primer on complicated calculus topics**

Calculus II is a prerequisite for lots of renowned collage majors, together with pre-med, engineering, and physics. *Calculus II For Dummies* deals professional guideline, suggestion, and the best way to support moment semester calculus scholars get a deal with at the topic and ace their exams.

It covers intermediate calculus subject matters in undeniable English, that includes in-depth insurance of integration, together with substitution, integration ideas and while to exploit them, approximate integration, and flawed integrals. This hands-on consultant additionally covers sequences and sequence, with introductions to multivariable calculus, differential equations, and numerical research. better of all, it comprises functional workouts designed to simplify and improve knowing of this advanced subject.

- Introduction to integration
- Indefinite integrals
- Intermediate Integration topics
- Infinite series
- Advanced topics
- Practice exercises

Confounded by means of curves? at a loss for words by way of polynomials? This plain-English advisor to Calculus II will set you straight!

**Preview of Calculus II For Dummies PDF**

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**Additional info for Calculus II For Dummies**

16x 2 - 1 three. convey those items by way of trig features of θ. within the secant case (as within the tangent case), all trig services may be in the beginning represented as tangents and secants. to symbolize the unconventional element as a trig functionality of θ, construct a fragment by utilizing the unconventional 16x 2 - 1 because the numerator and the consistent 1 because the denominator. Then set this fraction equivalent to the best trig functionality: 16x 2 - 1 = tan θ 1 discover that this fraction is the other facet of the triangle over the adjoining part c O m , so it equals tan θ. Simplifying it a piece grants this A equation: 1 = 1 16x 2 - 1 tan θ subsequent, convey dx as a trig functionality of θ. to take action, construct one other fraction with the variable x within the numerator and the consistent 1 within the denominator: 4x = sec θ 1 This time, the fraction is the hypotenuse over the adjoining facet of the triangle c H m , which equals sec θ. Now, remedy for x and differentiate to A locate dx: x = 1 sec θ four dx = 1 sec θ tan θ dθ four four. exhibit the necessary by way of θ and review it: # 1 dx = 16x 2 - 1 =1 four # sec θ d θ # tan1 θ $ 14 sec θ tan θ d θ Now, use the formulation for the crucial of the secant functionality from “Integrating the Six Trig capabilities” past during this bankruptcy: = 1 ln |sec θ + tan θ| + C four Chapter 7: Trig Substitution: understanding the entire (Tri)Angles five. swap the 2 θ phrases again into x phrases: for that reason, you don’t need to locate the worth of θ since you already comprehend the values of sec θ and tan θ when it comes to x from Step three. So, replacement those values to get your ultimate solution: = 1 ln 4x + 16x 2 - 1 + C four realizing whilst to prevent trig substitution Now that you simply know the way to take advantage of trig substitution, I offer you a ability that may be much more helpful: heading off trig substitution in case you don’t want it. for instance, examine the subsequent vital: # _1 - 4x i dx 2 2 this may seem like a superb position to take advantage of trig substitution, yet it’s a good greater position to take advantage of a bit algebra to extend the matter right into a polynomial: = # _1 - 8x 2 + 16x four i dx equally, examine this quintessential: # x dx x 2 - forty nine you should use trig substitution to guage this fundamental in an effort to. (You may also stroll to the pinnacle of the Empire nation development rather than taking the elevator if that tickles your fancy. ) besides the fact that, the presence of that little x within the numerator should still tip you off that variable substitution will paintings simply in addition (flip to bankruptcy five for extra on variable substitution): permit u = x2 – forty nine du = 2x dx 1 2 du = x dx utilizing this substitution leads to the subsequent critical: =1 2 # 1 du u = u +C = x 2 - forty nine + C performed! I most likely don’t have to inform you how a lot time and aggravation it can save you by way of operating smarter instead of tougher. So I won’t! 171 172 half II: Indefinite Integrals Chapter eight whilst All Else Fails: Integration with Partial Fractions during this bankruptcy ᮣ Rewriting complex fractions because the sum of 2 or extra partial fractions ᮣ figuring out find out how to use partial fractions in 4 exact instances ᮣ Integrating with partial fractions ᮣ utilizing partial fractions with flawed rational expressions L et’s face it: At this element on your math occupation, you've got higher issues to fret approximately than including a few fractions.