Trig Integral sin^2 x + cos^2 x = 1

In summary: That's correct.In summary, when solving the integral \intsin^{5}x cos x dx, you can use the trig identity sin^{2}x + cos^{2} x = 1 to rewrite sin^{5}x as (sin^{2}x)^{2} sin x. Then, using a u-substitution of u = cos x, you can integrate with respect to u and obtain the correct answer of -\frac{1}{6}cos^{6}x + \frac{1}{2}cos^{4}x - \frac{1}{2}cos^{2}x + C. Alternatively, you can also make a u-substitution directly on the initial problem and use the identity
  • #1
Youngster
38
0

Homework Statement



[itex]\int[/itex]sin[itex]^{5}[/itex]x cos x dx

Homework Equations



sin[itex]^{2}[/itex]x + cos[itex]^{2}[/itex] x = 1

The Attempt at a Solution



I've at least written down that sin[itex]^{5}[/itex]x = (sin[itex]^{2}[/itex]x)[itex]^{2}[/itex] sin x. Then I set sin[itex]^{2}[/itex]x equal to 1 - cos[itex]^{2}[/itex]x.

I then did a u-substitution, setting u equal to cos x to remove the sin x, and preceeded to integrate with respect to u. I ended up with [itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

The answer I received was [itex]\frac{1}{6}[/itex]sin[itex]^{6}[/itex]x + C

Can I get some insight on how to obtain that?

My textbook does give a procedure for this specific case (where one of the powers is odd), but I don't get the following instructions: "Then we combine the single sin x with dx in the integral and set sinxdx equal to -d(cosx)" which is stated after using the identity to replace the (sin[itex]^{2}[/itex]x)
 
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  • #2
Youngster said:

Homework Statement



[itex]\int[/itex]sin[itex]^{5}[/itex]x cos x dx

Homework Equations



sin[itex]^{2}[/itex]x + cos[itex]^{2}[/itex] x = 1

The Attempt at a Solution



I've at least written down that sin[itex]^{5}[/itex]x = (sin[itex]^{2}[/itex]x)[itex]^{2}[/itex] sin x. Then I set sin[itex]^{2}[/itex]x equal to 1 - cos[itex]^{2}[/itex]x.

I then did a u-substitution, setting u equal to cos x to remove the sin x, and preceeded to integrate with respect to u. I ended up with [itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

The answer I received was [itex]\frac{1}{6}[/itex]sin[itex]^{6}[/itex]x + C

Can I get some insight on how to obtain that?

My textbook does give a procedure for this specific case (where one of the powers is odd), but I don't get the following instructions: "Then we combine the single sin x with dx in the integral and set sinxdx equal to -d(cosx)" which is stated after using the identity to replace the (sin[itex]^{2}[/itex]x)
Do a u substitution directly on the problem as it was initially given to you.

What is the derivative of sin(x) ?
 
  • #3
...

Right, I feel kinda "slow" now.

Any way to use the trig identity to solve this though?
 
  • #4
... Make the substitution u = sinx and find du just like sammy said. The rest will follow.
 
  • #5
Zondrina said:
... Make the substitution u = sinx and find du just like sammy said. The rest will follow.

Yeah, I did that. Just curious if I can still use the identity to solve this though. Otherwise, I suppose that's it
 
  • #6
Youngster said:
Yeah, I did that. Just curious if I can still use the identity to solve this though. Otherwise, I suppose that's it

Yes, you can do it that way as well. Though I think there's a sign error in the answer you gave.
 
  • #7
-[itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

I think I was missing that negative. Is this correct?
 
  • #8
Youngster said:
-[itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

I think I was missing that negative. Is this correct?

Yep.
 

What is a trigonometric integral?

A trigonometric integral is an integration problem that involves trigonometric functions such as sine, cosine, and tangent. These integrals can be solved using various techniques, including substitution and trigonometric identities.

Why are trigonometric integrals important?

Trigonometric integrals are important because they are used in many real-world applications, such as physics, engineering, and mathematics. They also play a crucial role in solving differential equations, which are used to model a wide range of natural phenomena.

How do you solve a trigonometric integral?

There are various techniques for solving trigonometric integrals, including substitution, integration by parts, and using trigonometric identities. It is important to first identify the type of integral and then choose the appropriate method for solving it.

What are some common trigonometric identities used in solving integrals?

Some common trigonometric identities used in solving integrals include the Pythagorean identities, double angle identities, and half angle identities. These identities can be used to simplify the integral and make it easier to solve.

Are there any special cases in trigonometric integrals?

Yes, there are some special cases in trigonometric integrals that require specific techniques to solve. These include integrals involving inverse trigonometric functions and integrals with trigonometric functions raised to a power.

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