Learning Target 8: I can evaluate integrals involving products and powers of trigonometric functions. I can use trigonometric identities to solve integrals after making use of an appropriate trigonometric substitution. I can integrate rational functions by applying the method of partial fraction decomposition, including cases in which the decomposition involves repeated factors and irreducible quadratic factors. Notes: • If doing a trigonometric substitution, you must clearly indicate the substitution you make. If you are undoing a substitution, you must show the properly-labelled triangle used to write your answer in terms of the original variable of integration. Any compositions of a trig and an inverse trig function must be written in algebraic form. • If doing a partial fraction decomposition, you must show the work you have done by hand to determine the decomposition constants. You may check your work using Geogebra or WolframAlpha. Simplify your final answers. • Obviously, you must incorporate the methods of trigonometric substitution and/or the method of partial fraction decomposition to evaluate each integral. 1. Evaluate ? ?(x - 2) / ?(x - 1) dx Note (for question 1): What is the domain of the integrand? Start by making the substitution u² = x - 1 Does this substitution impose a restriction on the domain of the original integrand that does not already exist? After making the substitution, proceed with the method of trigonometric substitution.
Added by Tyler R.
Close
Step 1
Step 1: Find the domain of the integral by setting the denominator equal to 0: \( \sqrt{x} - 1 = 0 \) \( x = 1 \) So, the domain of the integral is \( x \neq 1 \). Show more…
Show all steps
Your feedback will help us improve your experience
Supreeta N and 61 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
trigonometry substitution
Supreeta N.
Place help me out with this trig substitution I can not seem to get to the right answer, Thank you
Evaluate the integrals. Remember to include a constant of integration with the indefinite integrals. Your answers may appear different from those in the Answers section but may still be correct. For example, evaluating $I=\int \sin x \cos x d x$ using the substitution $u=\sin x$ leads to $I=\frac{1}{2} \sin ^{2} x+C ;$ using $u=\cos x$ leads to $I=-\frac{1}{2} \cos ^{2} x+C ;$ and rewriting $I=\frac{1}{2} \int \sin (2 x) d x$ leads to $I=-\frac{1}{4} \cos (2 x)+C .$ These answers are all equal except for different choices for the constant of integration $C: \frac{1}{2} \sin ^{2} x=-\frac{1}{2} \cos ^{2}+\frac{1}{2}=-\frac{1}{4} \cos (2 x)+\frac{1}{4}$ You can always check your own answer to an indefinite integral by differentiating it to get back to the integrand. This is often easier than comparing your answer with the answer in the back of the book. You may find integrals that you can't do, but you should not make mistakes in those you can do because the answer is so easily checked. (This is a good thing to remember during tests and exams.) $$\int_{\frac{\pi^{2}}{16}}^{\frac{\pi^{2}}{9}} \frac{2^{\sin \sqrt{x}} \cos \sqrt{x}}{\sqrt{x}} d x$$
Integration
The Method of Substitution
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD