MATH 156 LAB 9Topic 1: An interesting integral.We will consider the integral NiMtJSRpbnRHNiQqKiUkc2luRyIiIiUieEdGKCUkY29zR0YoRilGKEYp. We can substitute NiMvJSJ1RyomJSRzaW5HIiIiJSJ4R0Yn, which gives NiMvJSNkdUcqKCUkY29zRyIiIiUieEdGJyUjZHhHRic=. Consequently the integral is NiMtJSRpbnRHNiQlInVHRiY=. This gives NiMvKiYlInVHIiIjRiYhIiIqKCUkc2luR0YmJSJ4RyIiIkYmRic=.Introduce commands that ask Maple to compute this integral and verify the answer given.

Perform this substitution with Maple and show this result.

However, we could have used the substitution NiMvJSJ1RyomJSRjb3NHIiIiJSJ4R0Yn, which gives NiMvJSNkdUcsJCooJSRzaW5HIiIiJSJ4R0YoJSNkeEdGKCEiIg==. This gives NiMtJSRpbnRHNiQsJCUidUchIiJGJw==. This gives NiMvLCQqJiUidUciIiNGJyEiIkYoLCQqKCUkY29zR0YnJSJ4RyIiIkYnRihGKA==. As you see this answer is not the same as above. Perform this substitution with Maple and get this result.

Somehow the answers should match. It is NOT TRUE that NiMvKiYlJHNpbkciIiMlInhHIiIiLCQqJiUkY29zR0YmRidGKCEiIg==. What makes the difference is that we forgot the constants of integration. So the two answers differ by a constant. To see this we ask Maple to compute the difference and simplify. Do this.

There is another way to perform the integration: We can use the trigonometric identity NiMvLSUkc2luRzYjKiYiIiMiIiIlInhHRikqLEYoRilGJUYpRipGKSUkY29zR0YpRipGKQ==. Compute the integral using this identity and show that your answer matches with the previous two.

Topic 2: The method of partial fractions and completing the square.Quite often we have to integrate functions that are quotients of two polynomials NiMqJi0lIlBHNiMlInhHIiIiLSUiUUdGJiEiIg==. These functions are called rational functions and we can use the method of partial fractions to integrate them. Example:NiMtJSRpbnRHNiQqJiwoKiQlInhHIiIkIiIiKiYiIiNGKyokRilGLUYrISIiIiIoRitGKywsKiRGKSIiJUYrKiZGKkYrRihGK0YvKiZGKkYrRi5GK0YrKiZGKkYrRilGK0YvRi1GK0YvRik= .We introduce the expressionf:=(x^3-2*x^2+7)/(x^4-3*x^3+3*x^2-3*x+2);

NiM+JSJmRyomLCgqJCklInhHIiIkIiIiRisqJiIiI0YrKUYpRi1GKyEiIiIiKEYrRissLCokKUYpIiIlRitGKyomRipGK0YoRitGLyomRipGK0YuRitGKyomRipGK0YpRitGL0YtRitGLw==

fpar:=convert(f, parfrac,x);

NiM+JSVmcGFyRywoKigiIiYhIiIsJiIiJyIiIiomIiM4RislInhHRitGK0YrLCYqJClGLiIiI0YrRitGK0YrRihGKyooIiIoRitGJ0YoLCZGLkYrRjJGKEYoRisqJiIiJEYrLCZGLkYrRitGKEYoRig=

fpar is equal to f, only it is written in a form that is easier to integrate. Now we can define the integral of fpar and ask Maple to compute it. We recognize ourselves that NiMsJiUieEciIiIiIiMhIiI= in the denominator will give NiMtJSNsbkc2IywmJSJ4RyIiIiIiIyEiIg== and the NiMsJiUieEciIiJGJSEiIg== in the denominator will give NiMtJSNsbkc2IywmJSJ4RyIiIkYoISIi. Also we can split the expression with NiMsJiokJSJ4RyIiIyIiIkYnRic= in the denominator into NiMqJiIiJyIiIi0iIiY2IywmKiQlInhHIiIjRiVGJUYlISIi plus NiMqKCIjOCIiIiUieEdGJS0iIiY2IywmKiRGJiIiI0YlRiVGJSEiIg==. The first term gives an arctan (NiMlInhH), while the second gives NiMtJSNsbkc2IywmKiQlInhHIiIjIiIiRipGKg== with appropriate constant in front.fint:=Int(fpar, x);

fvalue:=value(fint);

Naturally Maple can do all these things by itself:gint:=Int(f, x);

Find the partial fraction decomposition and integrate:NiMtJSRpbnRHNiQqJiwoKiQlInhHIiIpIiIiKiYiIiNGK0YpRitGK0YrISIiRisqJiwmRilGK0YrRi4iIiQsJiokRilGLUYrRjFGK0YtRi5GKQ== .

Sometimes this is not working because the roots of the denominator are not real numbers, or involve radicals. If the denominator does not have real roots, we can complete the square. The command is completesquare(expression, x) .

f:=1/(x^2+6*x+14);

fpar:=convert(f, parfrac, x);

newf:=completesquare(f, x);

This suggests the substitution NiMvJSJ1RywmJSJ4RyIiIiIiJEYn. Perform this substitution and compute the integral.

Naturally Maple can do all these intermediate steps at once. Write the corresponding command.

Compute the integral NiMtJSRpbnRHNiQqJiwqKiQlInhHIiIkIiIiKiYiIiZGKyokRikiIiNGK0YrKiYiIihGK0YpRishIiJGK0YrRissKEYuRitGKUYrRitGK0YyRik=.

Verify the answer by doing the integration by hand.