# Area between two trigonometric curves.

#### alane1994

##### Active member
http://www.mathhelpboards.com/f12/im-clueless-how-start-2322/[HR][/HR]My math class uses an online homework system. I got the answer wrong to the question, but I can get a similar question. Here is one that is similar to the earlier one.

You have two functions that are graphed.
$$y=\frac{\csc^2{x}}{4}$$
$$y=4\sin^2{x}$$

The purpose of the problem is to find the area between the curves.

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#### Jameson

Staff member
Re: Related to clueless

1) If you bring up that this is homework you also should state your professor is ok with you receiving guidance as long as you aren't just getting final answers without any effort. If you don't express this it could come off as cheating and waste time explaining this to the moderators after the fact.

2) What have you tried so far? This is going to be an integral and from the other thread we've already established the bounds, so show us what you've done.

This is not meant to be rude, just some advice on how you can get help the fastest and in the most efficient way for everyone

#### alane1994

##### Active member
Re: Related to clueless

1) OK, my professor is OK with me getting help as long as I am not just getting answers.
2) I have done the roots so far... I believe they are
$$x=\frac{\pi}{6},\frac{5\pi}{6}$$
Although, I am not too confident in this answer...

#### alane1994

##### Active member
Re: Related to clueless

Then
$$\text{Area}=\int^b_a[f(x)-g(x)]dx$$

$$\text{Area}=\int^\frac{5\pi}{6}_\frac{\pi}{6}[4\sin^2{x}-\frac{\csc^2{x}}{4}]dx$$

#### Jameson

Staff member
Re: Related to clueless

Again we need a domain because of these two functions will keep intersecting but assuming the question is looking for the area of one of these regions then one pair of lower and upper bounds is indeed $$\displaystyle \left[ \frac{\pi}{6},\frac{5\pi}{6} \right]$$. Your integral looks good. If you're doing it by hand then this calculation has a lot of room for mistakes. Walk us through your work on the integral now.