# Thread: Definite and Indefinite Integrals

1. ## 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.  Reply With Quote

2. Originally Posted by ME_student 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........
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  Reply With Quote

3. Originally Posted by ME_student 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.
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)*(cos(x)^2)*cos(x). Work from there.

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.
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.  Reply With Quote

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

5. Originally Posted by topo 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)*(cos(x)^2)*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.
For problem one I see where I can go from there, thanks.

For problem 2 I think I got it figured out.  Reply With Quote

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