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>
Answer: 6 inches
Step-by-step explanation:
1.5 miles wide * 1 inch / 0.25 miles = 6 inches :D
First of all we need to realize that half of a yard is as much as 18 inches. If a diameter of the circle is 18 inches, then its radius is 18 ÷ 2 = 9 inches.
Answer:
20 seconds
Step-by-step explanation:
Your friend starts 10 meters in front of the starting line.
You start from the starting line.
You run at 6 meters/ second.
Your friend runs at 5.5 meters/second.
You and your friend will be at the same distance from the starting line only when you are able to catch your friend.
The relative speed between you two is the difference in your speed and your friend's speed that is:
So to cover 10 meters of separation at the speed of 0.5 meter/second it will take:
So it will take 20 seconds for you and your friend to be at the same distance from the starting line.
Answer:
i think its f(x)=-2(x-2)(by the power of 2)-2
Step-by-step explanation:
