Answer:
how can I tell u the answer which out a pic
Step-by-step explanation:
Answer:
y = 2
Step-by-step explanation:
Kindly view the attached image to see the rules when looking for the horizontal asymptote.
In this situation we have 2x² / x²
The powers of both are equal to each other therefore the horizontal asymptote will be at the coefficient of the numerator divided by the coefficient of the denominator.
In other words the horizontal asymptote is at y = 2/1 or just 2
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Evaluate the indefinite integral:

Trigonometric substitution:

then,
![\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcl%7D%20%5Cmathsf%7Bx%3Dsin%5C%2C%5Ctheta%7D%26%5Cquad%5CRightarrow%5Cquad%26%5Cmathsf%7Bdx%3Dcos%5C%2C%5Ctheta%5C%2Cd%5Ctheta%5Cqquad%5Ccheckmark%7D%5C%5C%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bx%5E2%3Dsin%5E2%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bx%5E2%3D1-cos%5E2%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bcos%5E2%5C%2C%5Ctheta%3D1-x%5E2%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bcos%5C%2C%5Ctheta%3D%5Csqrt%7B1-x%5E2%7D%5Cqquad%5Ccheckmark%7D%5C%5C%5C%5C%5C%5C%20%26%26%5Ctextsf%7Bbecause%20%7D%5Cmathsf%7Bcos%5C%2C%5Ctheta%7D%5Ctextsf%7B%20is%20positive%20for%20%7D%5Cmathsf%7B%5Ctheta%5Cin%20%5Cleft%5B%5Cdfrac%7B%5Cpi%7D%7B2%7D%2C%5C%2C%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D.%7D%20%5Cend%7Barray%7D)
So the integral

becomes

Integrate

by parts:


Substitute back for the variable x, and you get

I hope this helps. =)
Tags: <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>
X^3-1 rewrite as x^3-1^3
differences of cubes formula: a^3-b^3=(a-b)(a^2+ab+b^2)
a=x, b=1
(x-1)(x^2+x*1+1^2)
(x-1)(x^2+x+1)