integratation

Aruwin

Hi again,I have another problem to solve.
What is the integral of cos^3xcos3x dx from 0 to pi/2??

I tried to use:
cos^3x=(1-sin^2x)cosx
=cosx-sin^2xcosx

So now it becomes IN{0 to pi/2} (cosx-sin^2xcosx)cos3x dx
what should I do?Is there any less complicated way?

In integration, always take care that you convert the power terms(raise to or exponential terms) into linear terms if there is no derivative of function present.
Say its ((cosx)^2)*sinx, then there is no need of conversion, we can substitute but generally if you donot spot any relation where f(x)*f'(x) are in the same expression, go for making linear terms, it works.

Also inorder to solve such types of integration, it is important that you are well versed with trignometric relations.

If I were you, I would have used cos3x=4(cos x)^3-3(cosx).
Now as I mentioned before, convert them to linear terms.
Replace (cos x)^3= (cos3x+3cosx)/4.
Then you would get some term where you have (cos(3x))^2 simplify this as (1+cos6x)/2.(See we are interested in linear terms only.)
In the next part you would get a term in form of cosA*cosB replace them as ((cos(A+B))+(cos(A-B)))/2

Now you have done with this, all terms now are linear and hence you can integrate it easily.
I am purposely not giving the solution so that you would have to try. Still If you dont get then show your new work and I would write down the solution. The above method was using simple techniques.

The answer to this should be pi/16. If you get it, well and good, otherwise I would show you how.

Take care.

Aruwin

Hi again,I have another problem to solve.
What is the integral of cos^3xcos3x dx from 0 to pi/2??

I tried to use:
cos^3x=(1-sin^2x)cosx
=cosx-sin^2xcosx

So now it becomes IN{0 to pi/2} (cosx-sin^2xcosx)cos3x dx
what should I do?Is there any less complicated way?

In integration, always take care that you convert the power terms(raise to or exponential terms) into linear terms if there is no derivative of function present.
Say its ((cosx)^2)*sinx, then there is no need of conversion, we can substitute but generally if you donot spot any relation where f(x)*f'(x) are in the same expression, go for making linear terms, it works.

Also inorder to solve such types of integration, it is important that you are well versed with trignometric relations.

If I were you, I would have used cos3x=4(cos x)^3-3(cosx).
Now as I mentioned before, convert them to linear terms.
Replace (cos x)^3= (cos3x+3cosx)/4.
Then you would get some term where you have (cos(3x))^2 simplify this as (1+cos6x)/2.(See we are interested in linear terms only.)
In the next part you would get a term in form of cosA*cosB replace them as ((cos(A+B))+(cos(A-B)))/2

Now you have done with this, all terms now are linear and hence you can integrate it easily.
I am purposely not giving the solution so that you would have to try. Still If you dont get then show your new work and I would write down the solution. The above method was using simple techniques.

The answer to this should be pi/16. If you get it, well and good, otherwise I would show you how.

Take care.

simplycoder

In integration, always take care that you convert the power terms(raise to or exponential terms) into linear terms if there is no derivative of function present.
Say its ((cosx)^2)*sinx, then there is no need of conversion, we can substitute but generally if you donot spot any relation where f(x)*f'(x) are in the same expression, go for making linear terms, it works.

Also inorder to solve such types of integration, it is important that you are well versed with trignometric relations.

If I were you, I would have used cos3x=4(cos x)^3-3(cosx).
Now as I mentioned before, convert them to linear terms.
Replace (cos x)^3= (cos3x+3cosx)/4.
Then you would get some term where you have (cos(3x))^2 simplify this as (1+cos6x)/2.(See we are interested in linear terms only.)
In the next part you would get a term in form of cosA*cosB replace them as ((cos(A+B))+(cos(A-B)))/2

Now you have done with this, all terms now are linear and hence you can integrate it easily.
I am purposely not giving the solution so that you would have to try. Still If you dont get then show your new work and I would write down the solution. The above method was using simple techniques.

The answer to this should be pi/16. If you get it, well and good, otherwise I would show you how.

Take care.

Thank you so much for the hints.I have done it!!Well,just in case there is something wrong with my working,I am showing you the solution I did.

Cool..........
It just reminded me when I was in 12th...😀