Integration by Parts with sin and ln(x)

[Ayesh]

Homework Statement

The method to use to integrate the function is up to us.
The choices are:
1) U-substitution
2)Integration by Parts
3)Trigonometric integrals
4)Trigonometric substitution
5)Partial fraction

Homework Equations

According to me, the best way to do it is to use Integration by Parts.

Here is the function:

$$\int$$sin(ln(x))/x²

The Attempt at a Solution

u=sin(lnx)
du=cos(lnx)/x dx
dv=1/x² dx
v=-1/x

$$\int$$sin(lnx)/x² dx = -sin(lnx)/x - $$\int$$ -cos(lnx)/x² dx
= -sin(lnx)/x + $$\int$$ cos(lnx)/x² dx
= -sin(lnx)/x + (-cos(lnx)/x² - $$\int$$ sin(lnx)/x dx

2 $$\int$$ sin(lnx)/x² dx = -sin(lnx)/x - cos(lnx)/x² dx

$$\int$$sin(lnx)/x²dx = 1/2(-sin(lnx)/x - cos(lnx)/x²)



According to Maple, the answer is 1/2((-2sin(lnx) - 2cos(1/2ln(x))² + 1 + 2sin(1/2ln(x))cos(1/2lnx))/x

 

[tiny-tim]

Hi Ayesh!

(have an integral: ∫ and try using the X² tag just above the Reply box )

…Int sin(lnx)/x^2 dx = 1/2(-sin(lnx)/x - cos(lnx)/x^2)

No, almost but not quite …

(and I'm afraid what you've written is too difficult to read for me to work out where the error was

According to Maple, the answer is 1/2((-2sin(lnx) - 2cos(1/2ln(x))^2 + 1 + 2sin(1/2ln(x))cos(1/2lnx))/x

oh, this is just Maple going beserk …

if you simplify that, you'll get something much more sensible

 

[Ayesh]

The answer Maple gave me is simplified.

 

[tiny-tim]

The answer Maple gave me is simplified.

really?? simplify it some more!

 

[Ayesh]

This is the most simplified answer Maple gives me :

-(1/2)*(2sin(lnx)+2cos((1/2)ln(x))²-1-2sin((1/2)lnx)*cos((1/2)lnx))/x

 

[tiny-tim]

You're human! Maple isn't!

simplify it some more! ​

(alternatively, try the integration by parts again, only more neatly, and see if it comes out right this time )

 

[Ayesh]

I've redone the integration by hand and my answer now is:

$$\int$$sin(lnx)/x² = 1/2 (-sin(lnx) + cos(lnx)/x)

 

[vela]

Closer, but you flipped a sign somewhere.

You can always differentiate your answer and see if you recover the integrand to see if your answer is correct.

 

[tiny-tim]

Hi Ayesh!

(what happened to that ∫ i gave you? )

Closer, but you flipped a sign somewhere.

You can always differentiate your answer and see if you recover the integrand to see if your answer is correct.

I strongly agree.

Getting the sign wrong in integration by parts is notoriously easy,

so it's essential to check the answer by differentiating it (especially if the question has obviously been designed to put in as many minuses as possible!)

 

[Ayesh]

I found my mistake, thank you!

Differentiating was a good idea :)

 

[Count Iblis]

You could also have substituted x = exp(u) and then have done the partial integration. In that case there is an alternative method to partial integration that is not on your list:

6) Differentiation w.r.t. to a parameter.

If 6) works that's usually a lot easier than partial integration.