Integration by parts (problem plus question)

[dwdoyle8854]

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) ?

 

[Curious3141]

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.

 

[dwdoyle8854]

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

 

[Curious3141]

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

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.