We are asked in the problem to evaluate the integral of <span>(cosec^2 x-2005)÷cos^2005 x dx. The function is an example of a complex function with a degree that is greater than one and that uses special rules to integrate the function via the trigonometric functions. For example, we integrate
2005/cos^2005x dx which is equal to 2005 sec^2005 x since sec is the inverse of cos. The integral of this function when n >3 is equal to I=</span><span>∫<span>sec(n−2)</span>xdx+∫tanx<span>sec(n−3)</span>x(secxtanx)dx
Then,
</span><span>∫tanx<span>sec(<span>n−3)</span></span>x(secxtanx)dx=<span><span>tanx<span>sec(<span>n−2)</span></span>x/(</span><span>n−2)</span></span>−<span>1/(<span>n−2)I
we can then integrate the function by substituting n by 3.
On the first term csc^2 2005x / cos^2005 x we can use the trigonometric identity csc^2 x = 1 + cot^2 x to simplify the terms</span></span></span>
Example: Set up proportions.
3/4 cups → 1 batch
x cups → 2 batches
x = 2(3/4) = 6/4 = 1 1/2 cups
3/4 cups → 1 batch
y cups → 9 batches
y = 9(3/4) = 6 3/4 cups
Answer:
uhh
Step-by-step explanation:
a) 
has CDF
where the last equality follows from independence of 
. In terms of the distribution and density functions of , this is
Then the density is obtained by differentiating with respect to 
,
b) 
can be computed in the same way; it has CDF
Differentiating gives the associated PDF,
Assuming 
and 
, we have
and
I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.
