Definite and Indefinite Integrals

Hello,

I have two handout problem from my professor that I can't quite figure out on my own. These two problems are NOT homework problems.

So we have the integral of (sin(x))^2*(cos(x))^3 dx. The hint I was given was to rewrite it as the integral (sin(x))^2*(1-(sin(x))^2*(cos(x)) dx. Is that even allowed? There is no trig identity where I am allowed to do this... This problem is a Definite integral from a=pi/4 to b=pi/2.


The other problem I was having trouble with is to take the indefinite integral of arc sec(sqrx) dx. The hint my professor gave me was to let: u=arc sec(sqrx) and dv=dx. I think I am just having trouble with some algebra in this problem. If you gentlemen figure it out could you please post a picture of the problem worked out? Thank you.

[QUOTE=ME_student;5715]Hello,

I have two handout problem from my professor that I can't quite figure out on my own. These two problems are NOT homework problems........

[/QUOTE]

How is this not homework?

small hint ..for prob1

cos(x)dx=?

small hint for problem 2
Draw a right triangle and enter u and use trig functions

[QUOTE=ME_student;5715]Hello,

I have two handout problem from my professor that I can't quite figure out on my own. These two problems are NOT homework problems.

So we have the integral of (sin(x))^2*(cos(x))^3 dx. The hint I was given was to rewrite it as the integral (sin(x))^2*(1-(sin(x))^2*(cos(x)) dx. Is that even allowed? There is no trig identity where I am allowed to do this... This problem is a Definite integral from a=pi/4 to b=pi/2.[/QUOTE]

Yes, there is a very simple trig identity but you seem to be focusing too much on what's written and not what it actually means.

Note that (sin(x)^2)*(cos(x)^3) = (sin(x)^2)*[B](cos(x)^2)[/B]*cos(x). Work from there.

[quote]The other problem I was having trouble with is to take the indefinite integral of arc sec(sqrx) dx. The hint my professor gave me was to let: u=arc sec(sqrx) and dv=dx. I think I am just having trouble with some algebra in this problem. If you gentlemen figure it out could you please post a picture of the problem worked out? Thank you.[/quote]

This problem requires integration by parts and the following theorem:

f(x)=y
If g(y)=x (i.e. it is the inverse function of f(x)), then therefore by the chain rule:

g'(y) = 1 = y'*g'(y) which means that 1/y' = g'(y) which means 1/f'(x) = g'(y).

In other words, the derivative of an inverse function (such as arcsec) is equal to 1 divided by the derivative of the function (in this case the sec function).

Hope this helps.

They were practice handout problems. You would be surprised how much non-turned in homework I do when I am not in class...

[QUOTE=topo;5721]Yes, there is a very simple trig identity but you seem to be focusing too much on what's written and not what it actually means.

Note that (sin(x)^2)*(cos(x)^3) = (sin(x)^2)*[B](cos(x)^2)[/B]*cos(x). Work from there.



This problem requires integration by parts and the following theorem:

f(x)=y
If g(y)=x (i.e. it is the inverse function of f(x)), then therefore by the chain rule:

g'(y) = 1 = y'*g'(y) which means that 1/y' = g'(y) which means 1/f'(x) = g'(y).

In other words, the derivative of an inverse function (such as arcsec) is equal to 1 divided by the derivative of the function (in this case the sec function).

Hope this helps.[/QUOTE]

For problem one I see where I can go from there, thanks.

For problem 2 I think I got it figured out.