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:
36
Step-by-step explanation:
Answer:
The answer is <u>54.6</u>.
9.1(6)+8.7(7)=<u>54.6</u>+60.9
Step-by-step explanation:
Given:
9.1(6)+8.7(7)=_+60.9
Now, to solve it:
9.1(6)+8.7(7)=_+60.9
Let the _be x.
So, we remove parenthesis first that makes it multiply with the number next to it:
Now adding on L.H.S:
Subtracting on both sides by 60.9 we get:
Therefore, the answer is 54.6.
A. First move all to the left side of the equation(Normal form)
x^3 - 49x= 0
B. Factor out an x, which is the GCF (Factored form)
x(x^2 - 49) = 0
C. Find solutions by making each x piece equal to 0. The first part is just x=0 and the second part is just factoring the difference of squares and then solving.
x=0, x^2 - 49 =0
x=0, x + 7 = 0, x - 7 = 0
Therefore, the answers for Part C are:
x = 0, x = -7, x = 7
