1. ## Integration by parts

Hi all,
Is this method/solution correct?
thanks
John

Integrate by parts;

$\displaystyle x^2.sin2x$

$\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx$

$\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx$

$\displaystyle v= -\frac{cos2x}{2}+C$

$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int\frac{du}{dx}.dx+C$

Substituting

$\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.dx+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\-cos2x.dx+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\frac{sin2x}{2}+C$

$\displaystyle =\frac{-x^2cos2x}{2}+\frac{sin2x}{4}+C$

2. ## Re: Integration by parts

Originally Posted by celtic1234
Hi all,
Is this method/solution correct?
thanks
John

Integrate by parts;

$\displaystyle x^2.sin2x$

$\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx$

$\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx$

$\displaystyle v= -\frac{cos2x}{2}+C$

$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int\frac{du}{dx}.dx+C$

Substituting

$\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.dx+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\-cos2x.dx+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\frac{sin2x}{2}+C$

$\displaystyle =\frac{-x^2cos2x}{2}+\frac{sin2x}{4}+C$
You have the rule wrong. The rule is $\displaystyle \displaystyle \int{u\,dv} = u\,v - \int{v\,du}$ (or if you prefer the longer notation: $\displaystyle \displaystyle \int{u\,\frac{dv}{dx}\,dx} = u\,v - \int{v\,\frac{du}{dx}\,dx}$ ).

3. ## Re: Integration by parts

sorry
this is my proper solution;
can you tell me if i am following the correct method-is the solution correct?
thanks
John

Integrate by parts;

$\displaystyle x^2.sin2x$

$\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx$

$\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx$

$\displaystyle v= -\frac{cos2x}{2}+C$

$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int\frac{du}{dx}.dx+C$

Substituting

$\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.dx+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\-cos2x.dx.2x+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\frac{sin2x}{2}.2x+C$

$\displaystyle =\frac{-x^2cos2x}{2}+\frac{2xsin2x}{4}+C$

4. ## Re: Integration by parts

Originally Posted by celtic1234
sorry
this is my proper solution;
can you tell me if i am following the correct method-is the solution correct?
thanks
John

Integrate by parts;

$\displaystyle x^2.sin2x$

$\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx$
Good so far.
$\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx$

$\displaystyle v= -\frac{cos2x}{2}+C$
Absolutely fine to here.
$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int\frac{du}{dx}.dx+C$
And then you stray slightly. The rule is, simply put and imprecisely:

$\displaystyle \int u\dfrac{dv}{dx}=u\cdot{v}-\int v\dfrac{du}{dx}$

When I substitute, this becomes:

$\displaystyle \int x^2sin2x=x^2\cdot{-\frac{cos2x}{2}}-\int -\frac{cos2x}{2}\cdot 2x~ dx$

$\displaystyle \int x^2sin2x=-x^2\cdot{\frac{cos2x}{2}}+\int x\cdot cos2x~dx$

And I think you'll have to use parts again here, making this a particularly bothersome question.

5. ## Re: Integration by parts

Thanks
Is this it?

Integrate by parts;

$\displaystyle x^2.sin2x$

$\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx$

$\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx$

$\displaystyle v= -\frac{cos2x}{2}+C$

$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

Substituting

$\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.2x.dx+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C$

Need to solve the integral of 2x.Cos2x by parts:

$\displaystyle \int 2x.cos2x$

$\displaystyle let \ u=2x, then\ \frac {du}{dx}=2.dx$

$\displaystyle let \ \frac{dv}{dx}=cos2x, then\ v= \int (cos2x).dx$

$\displaystyle v= \frac{sin2x}{2}+C$

$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

Substitute back in;

$\displaystyle =2x.\frac{sin2x}{2}-\int\frac{sin2x}{2}.2.dx+C$

$\displaystyle =2x.\frac{sin2x}{2}-\frac{2}{2}\int sin2x.dx+C$

$\displaystyle =2x.\frac{sin2x}{2}-1.\frac{-cos2x}{2}+C$

$\displaystyle =2x.\frac{sin2x}{2}+\frac{cos2x}{2}+C$

$\displaystyle =x.sin2x+\frac{cos2x}{2}+C$

6. ## Re: Integration by parts

Originally Posted by Quacky
And then you stray slightly. The rule is, simply put and imprecisely:

$\displaystyle \int u\dfrac{dv}{dx}=u\cdot{v}-\int v\dfrac{du}{dx}$

When I substitute, this becomes:

$\displaystyle \int x^2sin2x=x^2\cdot{-\frac{cos2x}{2}}-\int -\frac{cos2x}{2}\cdot 2x~ dx$

$\displaystyle \int x^2sin2x=-x^2\cdot{\frac{cos2x}{2}}+\int x\cdot cos2x~dx$

And I think you'll have to use parts again here, making this a particularly bothersome question.
Originally Posted by celtic1234
Thanks
Is this it?

Integrate by parts;

$\displaystyle x^2.sin2x$

$\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx$

$\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx$

$\displaystyle v= -\frac{cos2x}{2}+C$

$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

Substituting

$\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.2x.dx+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C$

Need to solve the integral of 2x.Cos2x by parts:

$\displaystyle \int 2x.cos2x$

$\displaystyle let \ u=2x, then\ \frac {du}{dx}=2.dx$

$\displaystyle let \ \frac{dv}{dx}=cos2x, then\ v= \int (cos2x).dx$

$\displaystyle v= \frac{sin2x}{2}+C$
Good so far.

$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

Substitute back in;

$\displaystyle =2x.\frac{sin2x}{2}-\int\frac{sin2x}{2}.2.dx+C$
Yep
$\displaystyle =2x.\frac{sin2x}{2}-\frac{2}{2}\int sin2x.dx+C$

$\displaystyle =2x.\frac{sin2x}{2}-1.\frac{-cos2x}{2}+C$

$\displaystyle =2x.\frac{sin2x}{2}+\frac{cos2x}{2}+C$

$\displaystyle =x.sin2x+\frac{cos2x}{2}+C$
Yes, I think that's fine!
You made it difficult for yourself. When you get to this stage:

$\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.2x.dx+C$

You can cancel the 2s and rewrite, as I had:

$\displaystyle =-x^2\cdot{\frac{cos2x}{2}}+\int x\cdot cos2x~dx$

Which is easier to work with.

$\displaystyle \int x^2sin(2x)~dx$ ? You don't have to work anything else out. Use these two facts:

$\displaystyle \int x^2sin(2x)~dx=x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C$

$\displaystyle \int\2x.cos2x~dx=x.sin2x+\frac{cos2x}{2}$

7. ## Re: Integration by parts

If $\displaystyle \displaystyle u = x^2$ then EITHER $\displaystyle \displaystyle \frac{du}{dx} = 2x$ OR $\displaystyle \displaystyle du = 2x\,dx$, NOT $\displaystyle \displaystyle \frac{du}{dx} = 2x\,dx$.

8. ## Re: Integration by parts

Thanks all
this is my full solution:
Sorry for reposting it all but in case anyone else wanted to follow the full solution-(i am sure to confuse them otherwise)
Hopefully i did not make anymore mistakes!
thanks
John

Integrate by parts;

$\displaystyle x^2.sin2x$

$\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx$

$\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx$

$\displaystyle v= -\frac{cos2x}{2}+C$

$\displaystyle rule=\ \int u \frac{dv}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

Substituting

$\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.2x.dx+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C.......equa tion (1)$

Need to solve the integral of 2x.Cos2x by parts:

$\displaystyle \int 2x.cos2x$

$\displaystyle let \ u=2x, then\ du=2.dx$

$\displaystyle let \ \frac{dv}{dx}=cos2x, then\ v= \int (cos2x).dx$

$\displaystyle v= \frac{sin2x}{2}+C$

$\displaystyle rule=\ \int u \frac{dv}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

Substitute back in;

$\displaystyle =2x.\frac{sin2x}{2}-\int\frac{sin2x}{2}.2.dx+C$

$\displaystyle =2x.\frac{sin2x}{2}-\frac{2}{2}\int sin2x.dx+C$

$\displaystyle =2x.\frac{sin2x}{2}-1.\frac{-cos2x}{2}+C$

$\displaystyle =2x.\frac{sin2x}{2}+\frac{cos2x}{2}+C$

$\displaystyle =x.sin2x+\frac{cos2x}{2}+C$

Now substitute this integral back into equation (1)

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C.......equa tion (1)$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}[x.sin2x+\frac{cos2x}{2}]$

$\displaystyle =-x^2.\frac{cos2x}{2}+\frac{x.sin2x}{2}+\frac{cos2x} {4}$

9. ## Re: Integration by parts

Nice! Perfect solution.

10. ## Re: Integration by parts

thanks Quacky
appreciate the help
I am working on another one now that has an Lnx term in it-it seems to be repeating itself over and over-no thought there is a simple solution -i may well have to post about it though....keep the eyes peeled!
John

11. ## Re: Integration by parts

no doubt...not no thought? i am getting worse by the day...

12. ## Re: Integration by parts

Okay

Maybe this will help - for $\displaystyle \int ln(x)~dx$, let $\displaystyle \frac{dv}{dx}=1$ and $\displaystyle u=ln(x)$

13. ## Re: Integration by parts

Hi Quacky,
If the function is

$\displaystyle x^3.lnx$

it repeats and repeats.

Now if i make

$\displaystyle u= x^3 and \frac{dv}{dx}=lnx$

Then i need to integrate the $\displaystyle lnx$ term to get v.

however if integrate it by parts and make

$\displaystyle \ u=1$ and$\displaystyle \frac{dv}{dx}=lnx$

then it repeats over and over.

However if i make:

$\displaystyle u=lnx$ and$\displaystyle \frac{dv}{dx}=1$

then i can get a solution fairly easily.

Is it ok to rearrange the $\displaystyle lnx$ and the $\displaystyle 1.dx$ to suit ?

Do i need to post this as a different thread?

regards
John

14. ## Re: Integration by parts

Sorry to butt in, but, just in case a picture helps...

Originally Posted by celtic1234
Hi Quacky,
If the function is

$\displaystyle x^3.lnx$

it repeats and repeats.
Not so.

... where (key in spoiler) ...

Spoiler:

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,

... is lazy integration by parts, doing without u and v.

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15. ## Re: Integration by parts

I don't know if you followed that, but let $\displaystyle u=ln(x)$ because you can differentiate that to give $\displaystyle \frac{1}{x}$ which simplifies the further stages simply.

Let $\displaystyle \frac{dv}{dx}=x^3$

Follow this approach for all $\displaystyle ln(x)$ type questions.

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