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:
Well 14 2/3 is 14.66 so it would be B but yea it would fit.
(x+3)(x+3) = x^2 -3 + 3x
x^2 + 3x + 3x + 9 = x^2 + 3x - 3
x^2 + 6x + 9 = x^2 + 3x - 3
6x + 9 = 3x - 3
3x + 9 = -3
3x = -12
x = -4
Check:
(-4+3)^2 = (-4)^2 - 3(1+4)
(-1)^2 = 16 - 3(5)
1 = 16-15
1 = 1 :)
Answer:
i think its 7n/4
Step-by-step explanation:
