Integration by parts (problem plus question)

In summary, the conversation discusses a problem with getting the right answer multiplied by a constant instead of divided by it. The solution involves making a substitution and using integration by parts to evaluate the integral of cos(√x). The correct steps are 2∫Acos(A)dA = 2(Asin(A) - ∫sin(A)dA), and in terms of x, the final answer is 2(√x)(sin(√x) + 2cos(√x) + c).
  • #1
dwdoyle8854
16
0

Homework Statement


I've run into this problem a few times, where I get the right answer, but multiplied by a constant where I would have it divided by the constant or vice versa.

"First make a substitution and then use integration by parts to evaluate the integral"

∫cos(√x)dx


Homework Equations



∫udv = uv - ∫vdu

The Attempt at a Solution



let A = √x
dA = dx/(2√x)
2(√x)dA = dx
A2 = x

2∫Acos(A)dA

let u=A
du=dA
dv= cos(A)
v = sin A + C

2∫Acos(A)dA = Asin(A) - ∫sin(A)
= Asin(A) + cos(A)
so, then

∫Acos(A)dA = (Asin(A) +cos(A))/2 +C

this is wrong, it should be 2Asin(A) + 2cos(A) + C and I am not sure where exactly I can remedy this.

I think my problem might be with 2∫Acos(A)dA = Asin(A) - ∫sin(A)
since I have an integral I am evaluating by parts multiplied by a constant, does
2∫Acos(A)dA = Asin(A) - ∫sin(A) => 2∫Acos(A)dA = 2(Asin(A) - ∫sin(A)) ?

or more generally c∫udv = c(uv - ∫vdu) ?
 
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  • #2
dwdoyle8854 said:

Homework Statement


I've run into this problem a few times, where I get the right answer, but multiplied by a constant where I would have it divided by the constant or vice versa.

"First make a substitution and then use integration by parts to evaluate the integral"

∫cos(√x)dx

Homework Equations



∫udv = uv - ∫vdu

The Attempt at a Solution



let A = √x
dA = dx/(2√x)
2(√x)dA = dx
A2 = x

2∫Acos(A)dA

let u=A
du=dA
dv= cos(A)
v = sin A + C

2∫Acos(A)dA = Asin(A) - ∫sin(A)
You missed out the "dA" there at the end. But most importantly, you missed out the factor of 2 on the right hand side.

$$\int A\cos A dA = A\sin A - \int \sin A dA$$

and when you multiply by 2, you should do it throughout.

= Asin(A) + cos(A)
so, then

∫Acos(A)dA = (Asin(A) +cos(A))/2 +C

What?! Why? Something is wrong with your algebra. See what I wrote above.

2∫Acos(A)dA = 2(Asin(A) - ∫sin(A)) ?

or more generally c∫udv = c(uv - ∫vdu) ?

Yes, these two statements are correct (except you've missed out a "dA" again).

You don't have to bother with constants of integration at all until the final answer step. Even writing dv = cos A, hence v = sin A + c is not necessary.

You should express everything in terms of the original variable (x) in the final answer.
 
  • #3
Okay thank you. My problem was not distributing the constant from
2∫Acos(A)dA = Asin(A) - ∫sin(A)dA
this should be
2∫Acos(A)dA=2( Asin(A) - ∫sin(A)dA)
and in terms of x:

=2(√x)(sin(√x) + 2cos(√x )

yes.
Thanks bunches :approve:
 
  • #4
dwdoyle8854 said:
Okay thank you. My problem was not distributing the constant from
2∫Acos(A)dA = Asin(A) - ∫sin(A)dA
this should be
2∫Acos(A)dA=2( Asin(A) - ∫sin(A)dA)
and in terms of x:

=2(√x)(sin(√x) + 2cos(√x )

yes.
Thanks bunches :approve:

Your brackets are off - it should be ##2(\sqrt x \sin \sqrt x + \cos \sqrt x) + c##. Be careful about things like this, and don't forget your constant at the end. :smile:
 

FAQ: Integration by parts (problem plus question)

1. What is integration by parts?

Integration by parts is a technique in calculus that allows us to find the integral of a product of two functions. It is based on the product rule from differentiation and involves using a specific formula to simplify the integration process.

2. When should I use integration by parts?

Integration by parts is most useful when the integrand (the function being integrated) is a product of two functions, especially when one of the functions is difficult to integrate by other means (such as trigonometric or logarithmic functions).

3. How do I use integration by parts?

To use integration by parts, you need to identify the two functions in the integrand and choose one to differentiate and the other to integrate. Then, use the integration by parts formula (u dv = uv - ∫v du) to simplify the integral. It may take several iterations of this process to fully integrate the original function.

4. What is the integration by parts formula?

The integration by parts formula, also known as the "u-substitution" formula, is u dv = uv - ∫v du. In this formula, u and v represent the two functions in the integrand, with u being the one chosen to differentiate and v being the one chosen to integrate. ∫v du represents the integral of v with respect to u.

5. What are some common mistakes when using integration by parts?

Some common mistakes when using integration by parts include choosing the wrong functions to differentiate and integrate, forgetting to add the "constant of integration" when integrating, and not using the formula correctly (such as forgetting to take the derivative of u or integrating v incorrectly). It is important to double check your work and practice using the formula to avoid these mistakes.

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