How Do You Integrate \(\sin 6x + 3\cos 5x\) Using Substitution?

In summary, the conversation discusses finding the integral of sin 6x + 3cos 5x through the 'u' substitution method. The individual provides their attempted solution and asks for confirmation on their approach. Another individual confirms that they are on the right track and offers advice on avoiding confusion with different variables. The original individual expresses their lack of strong Algebra background and thanks the other for their help.
  • #1
duki
264
0

Homework Statement



Integrate [tex]\int\sin 6x + 3\cos 5x dx[/tex]

Homework Equations

The Attempt at a Solution



The way I was taught was the 'u' substitution method.
I know that [tex]\int\sin u du = -\cos u + C[/tex] and I know that [tex]\int\cos u du = \sin u + C[/tex]

Here's what I've done so far

[tex]\int\sin 6x + 3\cos 5x dx[/tex]
[tex]\int\sin 6x dx + 3\int\cos 5x dx[/tex]
u = 6x; u = 5x
du = 6dx; du = 5dx
du/6 = dx; du/5 = dx

[tex]\frac{1}{6}\int\sin u du + \frac{3}{5}\int\cos u du[/tex]

Am I on the right track?
Thanks!
 
Last edited:
Physics news on Phys.org
  • #2
Yep, everything you've done looks good so far. Just evaluate the integrals and sub the u values back in and you should be finished.
 
  • #3
Answer: [tex]\frac{-1}{6}\cos 6x + \frac{3}{5}\sin 6x + C[/tex] ?
 
  • #4
Your're OK so far...

Such is the glacial pace of my ancient home computer... Yes, your antiderivative is correct.
 
Last edited:
  • #5
duki said:
Answer: [tex]\frac{-1}{6}\cos 6x + \frac{3}{5}\sin 6x + C[/tex] ?

Close, not quite. Keep in mind that you have two separate u values for the two different integrals.

(Sometimes it's easier to use both u's and v's to avoid confusion)
 
  • #6
[tex]
\frac{-1}{6}\cos 6x + \frac{3}{5}\sin 5x + C
[/tex]
?
 
  • #7
Yep, that's the answer I got. It seems like you've got the calculus down!

Sometimes the algebra is all that will mess you up, lol.
 
  • #8
Yeah, I don't have as strong of an Algebra background as I should.
Thanks for the help
 

1. What is basic integration with sin and cos?

Basic integration with sin and cos is a mathematical process used to find the indefinite integral of functions involving the trigonometric functions sine and cosine. It is a fundamental concept in calculus that involves finding the antiderivative of a given function.

2. What are the key steps to integrate sin and cos functions?

The key steps to integrate sin and cos functions are as follows:

  1. Identify the function to be integrated.
  2. Apply any necessary trigonometric identities to simplify the function.
  3. Use the power rule to integrate the function.
  4. Check for any constants and add them to the final answer.

3. Why is basic integration with sin and cos important?

Basic integration with sin and cos is important because it allows us to find the area under a curve of a given function. This concept is essential in many fields, including physics, engineering, and economics, as it helps us to solve real-world problems involving rates of change and motion.

4. Can basic integration with sin and cos be used to solve definite integrals?

Yes, basic integration with sin and cos can be used to solve definite integrals. After finding the indefinite integral, we can then substitute the upper and lower limits of integration to find the definite integral. This allows us to calculate the exact area under a curve within a given interval.

5. What are some common mistakes to avoid when integrating sin and cos?

Some common mistakes to avoid when integrating sin and cos include forgetting to apply the power rule, not simplifying the function using trigonometric identities, and forgetting to add constants to the final answer. It is also essential to check for any mistakes in algebraic manipulation and to double-check the final answer using differentiation.

Similar threads

Integral using substitution x = -u
    • Calculus and Beyond Homework Help
Replies
12
Views
1K
Applying integration to math problems
    • Calculus and Beyond Homework Help
Replies
5
Views
786
Integrate [cosec(30°+x)-cosec(60°+x)] dx in terms of tan x
    • Calculus and Beyond Homework Help
Replies
3
Views
346
Integrals that keep me up at night
    • Calculus and Beyond Homework Help
Replies
22
Views
1K
U-Substitution in trig integral
    • Calculus and Beyond Homework Help
Replies
3
Views
1K
How Do You Solve the Differential Equation dy/dx = 1 - y^2?
    • Calculus and Beyond Homework Help
Replies
12
Views
1K
Why does dividing by ##\sin^2 x## solve the integral?
    • Calculus and Beyond Homework Help
Replies
27
Views
3K
Integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x)
    • Calculus and Beyond Homework Help
Replies
2
Views
1K
Use substitution to solve the definite integral
    • Calculus and Beyond Homework Help
Replies
9
Views
189
Integrate: ## \int \cos^m(x)\sin^n(x)~dx ##
    • Calculus and Beyond Homework Help
Replies
11
Views
1K

Hot Threads

Recent Insights

Back
Top