Integrate t.sin(t^2-1) using integration by parts | Math solution

[DryRun]

Homework Statement

Find $\int t\sin(t^2-1)$

The attempt at a solution
I've used the method of integration by parts.
Let U=t, then dU =1
Let dV=$sin(t^2-1)$, then $V=-\frac{cos(t^2-1)}{2t}$

Then, $\int t\sin(t^2-1)=t.-\frac{cos(t^2-1)}{2t}- \int -\frac{cos(t^2-1)}{2t}.1$

Again, i use the method of integration by parts to find $\int -\frac{cos(t^2-1)}{2t}.1$
Let U=$cos(t^2-1)$, then dU=$-\frac{sin(t^2-1)}{2t}$
Let dV=$\frac{-1}{2t}$, then V=$-\frac{ln2t}{2}$

Then, $\int -\frac{cos(t^2-1)}{2t}=cos(t^2-1).-\frac{ln2t}{2}-\int -\frac{ln2t}{2}.-\frac{sin(t^2-1)}{2t}$

The integration cycle is endless. I'm stuck.

Or, i could have also tried for $\int -\frac{cos(t^2-1)}{2t}$
Let $U=\frac{-1}{2t}$, then $dU=\frac{2}{t^2}$
Let $dV=cos(t^2-1)$, then $V=\frac{sin(t^2-1)}{2t}$

Then, $\int -\frac{cos(t^2-1)}{2t}=\frac{-1}{2t}.\frac{sin(t^2-1)}{2t}-\int \frac{sin(t^2-1)}{2t}.\frac{2}{t^2}$

But still, it looks like it's going to go on forever.

 

[MednataMiza]

Why not move the t into the differential. That way when you construct the equation you will have to find the derivative of the sine and then substitute...
I will work that example out in a few mins

 

[DryRun]

I didn't understand what you meant, but i probably tried it already in my first post above. Anyway, I'm still waiting for your attempt at the solution.

 

[ehild]

Homework Statement

Find $\int t\sin(t^2-1)$

The attempt at a solution
I've used the method of integration by parts.
Let U=t, then dU =1
Let dV=$sin(t^2-1)$, then $V=-\frac{cos(t^2-1)}{2t}$

That is wrong.

Use integration by substitution.ehild

 

[DryRun]

Hi ehild

A tentative substitution... Let $t=\sin\theta$. I tried it but it doesn't look "friendly".
Here is what i get: $\int \sin\theta\sin(-\cos^2\theta).\cos\theta\,.d\theta=\int \frac{\sin 2\theta}{2}.\sin(-\cos^2\theta)$
I don't think i should use t=tan (half-angle) here. So, no idea what substitution to use.

 

[ehild]

Hi ehild

A tentative substitution... Let $t=\sin\theta$. I tried it but it doesn't look "friendly".

Why do you think it makes things easier?

You have the composite function sin(t^2-1). It is multiplied by t which is almost the derivative of the inner function t^2-1. Choose the inner function as new variable.


ehild

 

[DryRun]

Let $u=t^2-1$

$t=\sqrt{u+1}$

$du=2t\,.dt$, $dt=\frac{1}{2\sqrt{u+1}}\,.du$

Then, i get the answer: $-\frac{\cos u}{2}=-\frac{\cos (t^2-1)}{2}$

But in Wolfram, i get this answer: $-\frac{\cos (1-t^2)}{2}$

I suppose this means that the two answers are equivalent to each other?

 

[ehild]

How are t^2-1 and 1-t^2 related?
What about the parity of the cosine function?

ehild

 

[DryRun]

How are t^2-1 and 1-t^2 related?
What about the parity of the cosine function?

ehild

$(t^2-1)=-(1-t^2)$

$\cos -(t^2-1)=\cos (t^2-1)$

Therefore, $-\frac{\cos (t^2-1)}{2}=-\frac{\cos -(1-t^2)}{2}=-\frac{\cos (1-t^2)}{2}$

Am i correct?

 

[ehild]

Well, you dropped a pair of parentheses. cos(-(t^2-1)) is correct.

Is everything clear now?

Do not forget that you can use substitution whenever you have have a factor in the integrand which is a constant times the derivative of an inner function.

ehld

 

[MednataMiza]

Sharks, sorry... Got back from shopping - had to unload everything ...
Now back to the problem: Indeed, this kind of substitution you wrote is the most efficient
I tried solving by moving under the differential, but it gets too complicated and takes too much time to solve :)
So, in conclusion - this is the way it should be done :)

 

[DryRun]

I understand it now. As usual, thanks for your help, ehild.

 

[pongo38]

The key to this is to use the double angle 2t. You need to convert the expression to be in terms of 2t instead of t squared.

 

[SammyS]

Homework Statement

Find $\int t\sin(t^2-1)$

In my opinion, some of the problems encountered in this thread arise from dropping the dt in the above integral, which should br written:

$\displaystyle \int t\sin(t^2-1)dt$​

That way the substitution u = t²-1

gives du=(2t)dt .

Upon rewriting the integral it becomes pretty obvious where to plug everything in.

$\displaystyle \int t\sin(t^2-1)dt=\frac{1}{2}\int \sin(t^2-1)(2t)dt$

$\displaystyle =\frac{1}{2}\int \sin(u)du$​

 

[ehild]

I understand it now. As usual, thanks for your help, ehild.

You are welcome (as usual ) but I hope next time you remember the possibility of substitution.


ehild

 

[ehild]

In my opinion, some of the problems encountered in this thread arise from dropping the dt in the above integral, which should br written:

$\displaystyle \int t\sin(t^2-1)dt$​

You are right, Sammy, (as always ) dt must not be omitted.

ehild