# Integration Example: Sines an Cosines

Q:  ∫cos5(x)sin4(x)dx

A: Some books have tables and charts to memorize on how to integrate these types of problems: what to do if the power is odd on one but even on the other, etc (called reduction formulas)… Boring… Who has time, space or desire to memorize formulas? Let’s solve the damn problem.

First, I see that there are sines and cosines, and we know that one is the derivative of the other (more or less).  This tells me that u-substitution is likely going to come up.

Remember: whenever you see a function and its derivative present in a problem, you want to be thinking u-substitution!

I also know that by using the Pythagorean Identity, sin2(x)+cos2(x)=1, I can convert an “even number” of cosines to sines and vice versa.

So, now that I know u-substitution is most likely, I want to leave behind one function to be “du” and the rest should be converted to “u’s”.

Cosine is literally the “odd man out”.  There is an odd number of cosines, so I will leave one cosine behind to eventually serve as du and convert the rest like so:

∫cos5(x)sin4(x)dx

∫cos(x)*cos4(x)*sin4(x)dx

∫cos(x)*(cos2(x))2*sin4(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx

Now, I can do a fairly clean u-substitution:

Let u = sin(x)

Then, du = cos(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx = ∫ du*(1-u2)2*u4

Or, rewrite it: ∫(1-u2)2*u4du

Simplify and clean up:

∫(1-2u2+u4)*u4du

∫u4-2u6+u8du

Integrate to get:

(1/5)u5-(2/7)u7+(1/9)u9+C

And, substitute back in u = sin(x):

(1/5)sin5(x)-(2/7)sin7(x)+(1/9)sin9(x)+C

This is only 1 way to solve this problem.  I could’ve picked a different u-substitution and I could’ve likely picked different trig identities to help me along the way.

# Multiplying Polynomials

Q: Multiply -9m(2m2 + 6m – 1)

A:  When adding and subtracting polynomials, you can only combined like terms.  This is not the case with multiplication.  You can multiply unlike terms together.

-9m(2m2 + 6m – 1)

In this problem the whole quantity in the parentheses is being multiplied by -9m.  So, -9m needs to multiply each term:

-9m(2m2 + 6m – 1) = -9m*2m2 + -9m*6m + -9m*-1

# Subtracting Polynomials

1:  Subtract: (-4y2-3y+8) – (2y2-6y+2)

A:  Here we are just asked to clean things up and combine like terms: basically, add apples with apples and oranges with oranges, but don’t accidentally add apples with oranges!

y2 terms get combined with y2 terms
y terms get combined with y terms
constants (lone numbers) get combined with constants

So, let’s use color to highlight the like terms:

(-4y2-3y+8) – (2y2-6y+2)

How many total y2 terms do you have? -4y2 – 2y2 = -6y2

How many total y terms do you have? -3y – -6y = -3y+6y = 3y

How many total constant terms do you have? 8+2 = 8 – 2 = 6

Did we forget anyone?  Did anyone not have a match that we need to bring along?  Nope! Everyone from both sets of parentheses is accounted for!

(-4y2-3y+8) – (2y2-6y+2) = -6y2 + 3y + 6

# Get Math Help from Me

My name is Stacey and I live in Corvallis, Oregon with my two kids.  I am passionate about math tutoring. Here are ways you can get help from me:

• I can tutor you on Skype — if you learn well from me, you can hire me.
• Browse through examples and useful tips here, or post a new question for me to answer.

You can contact me via Twitter or Facebook if you are interested in setting up a tutoring session:

# Use the area of a cirle to find the diameter

Q:  The area of a circle is 16(pi) cm squared.  What is the diameter of the circle?

First we need to know the area formula for a circle… This is:

A = pi*r^2  [r is the radius]

Since the area is given to us, we can plug that in to the formula to get:

16*pi = pi*r^2

The only variable is r, so we can use algebra to solve for r like so:

Divide both sides by pi (which makes the pi’s cancel out):

16*pi /pi = pi*r^2 /pi

16 = r^2

Now, take the square root of both sides to solve for r:

sqrt(16) = r

4 = r

So, the radius of the circle is 4.  The diameter is 2*radius… So, the diameter is 8 cm.
The text passage contains multiple math problems, but for the purpose of this task, I will focus on the first problem presented, which is the integration example.

Content structure: The passage presents a question involving the integration of a trigonometric function, followed by a detailed step-by-step solution.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Integration Example: Sines an Cosines

Q:  ∫cos5(x)sin4(x)dx

A: Some books have tables and charts to memorize on how to integrate these types of problems: what to do if the power is odd on one but even on the other, etc (called reduction formulas)… Boring… Who has time, space or desire to memorize formulas? Let’s solve the damn problem.

First, I see that there are sines and cosines, and we know that one is the derivative of the other (more or less).  This tells me that u-substitution is likely going to come up.

Remember: whenever you see a function and its derivative present in a problem, you want to be thinking u-substitution!

I also know that by using the Pythagorean Identity, sin2(x)+cos2(x)=1, I can convert an “even number” of cosines to sines and vice versa.

So, now that I know u-substitution is most likely, I want to leave behind one function to be “du” and the rest should be converted to “u’s”.

Cosine is literally the “odd man out”.  There is an odd number of cosines, so I will leave one cosine behind to eventually serve as du and convert the rest like so:

∫cos5(x)sin4(x)dx

∫cos(x)*cos4(x)*sin4(x)dx

∫cos(x)*(cos2(x))2*sin4(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx

Now, I can do a fairly clean u-substitution:

Let u = sin(x)

Then, du = cos(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx = ∫ du*(1-u2)2*u4

Or, rewrite it: ∫(1-u2)2*u4du

Simplify and clean up:

∫(1-2u2+u4)*u4du

∫u4-2u6+u8du

Integrate to get:

(1/5)u5-(2/7)u7+(1/9)u9+C

And, substitute back in u = sin(x):

(1/5)sin5(x)-(2/7)sin7(x)+(1/9)sin9(x)+C

This is only 1 way to solve this problem.  I could’ve picked a different u-substitution and I could’ve likely picked different trig identities to help me along the way.

# Multiplying Polynomials

Q: Multiply -9m(2m2 + 6m – 1)

A:  When adding and subtracting polynomials, you can only combined like terms.  This is not the case with multiplication.  You can multiply unlike terms together.

-9m(2m2 + 6m – 1)

In this problem the whole quantity in the parentheses is being multiplied by -9m.  So, -9m needs to multiply each term:

-9m(2m2 + 6m – 1) = -9m*2m2 + -9m*6m + -9m*-1

# Subtracting Polynomials

1:  Subtract: (-4y2-3y+8) – (2y2-6y+2)

A:  Here we are just asked to clean things up and combine like terms: basically, add apples with apples and oranges with oranges, but don’t accidentally add apples with oranges!

y2 terms get combined with y2 terms
y terms get combined with y terms
constants (lone numbers) get combined with constants

So, let’s use color to highlight the like terms:

(-4y2-3y+8) – (2y2-6y+2)

How many total y2 terms do you have? -4y2 – 2y2 = -6y2

How many total y terms do you have? -3y – -6y = -3y+6y = 3y

How many total constant terms do you have? 8+2 = 8 – 2 = 6

Did we forget anyone?  Did anyone not have a match that we need to bring along?  Nope! Everyone from both sets of parentheses is accounted for!

(-4y2-3y+8) – (2y2-6y+2) = -6y2 + 3y + 6

# Get Math Help from Me

My name is Stacey and I live in Corvallis, Oregon with my two kids.  I am passionate about math tutoring. Here are ways you can get help from me:

• I can tutor you on Skype — if you learn well from me, you can hire me.
• Browse through examples and useful tips here, or post a new question for me to answer.

You can contact me via Twitter or Facebook if you are interested in setting up a tutoring session:

# Use the area of a cirle to find the diameter

Q:  The area of a circle is 16(pi) cm squared.  What is the diameter of the circle?

First we need to know the area formula for a circle… This is:

A = pi*r^2  [r is the radius]

Since the area is given to us, we can plug that in to the formula to get:

16*pi = pi*r^2

The only variable is r, so we can use algebra to solve for r like so:

Divide both sides by pi (which makes the pi’s cancel out):

16*pi /pi = pi*r^2 /pi

16 = r^2

Now, take the square root of both sides to solve for r:

sqrt(16) = r

4 = r

So, the radius of the circle is 4.  The diameter is 2*radius… So, the diameter is 8 cm.
user
The text passage contains multiple math problems, but for the purpose of this task, I will focus on the first problem presented, which is the integration example.

Content structure: The passage presents a question involving the integration of a trigonometric function, followed by a detailed step-by-step solution.

TYPE: SINGLE_QUESTION_WITH_ANSWER