Answer:i need more info to answer this question
Step-by-step explanation:
Step 1: Factor both the numerator and denominator of the fraction. Step 2: Reduce the fraction. Step 3: Rewrite any remaining expressions in the numerator and denominator. Step 1: Factor both the numerator and denominator of the fraction.
Answer: f(x) = ∛(x+4)^2=8 + 3 = 11
4x5=20-9=11
Step-by-step explanation:
Answer:
x=7
Step-by-step explanation:
Answer:
The result of the integral is:
![\arcsin{(\frac{x}{3})} + C](https://tex.z-dn.net/?f=%5Carcsin%7B%28%5Cfrac%7Bx%7D%7B3%7D%29%7D%20%2B%20C)
Step-by-step explanation:
We are given the following integral:
![\int \frac{dx}{\sqrt{9-x^2}}](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7B9-x%5E2%7D%7D)
Trigonometric substitution:
We have the term in the following format:
, in which a = 3.
In this case, the substitution is given by:
![x = a\sin{\theta}](https://tex.z-dn.net/?f=x%20%3D%20a%5Csin%7B%5Ctheta%7D)
So
![dx = a\cos{\theta}d\theta](https://tex.z-dn.net/?f=dx%20%3D%20a%5Ccos%7B%5Ctheta%7Dd%5Ctheta)
In this question:
![a = 3](https://tex.z-dn.net/?f=a%20%3D%203)
![x = 3\sin{\theta}](https://tex.z-dn.net/?f=x%20%3D%203%5Csin%7B%5Ctheta%7D)
![dx = 3\cos{\theta}d\theta](https://tex.z-dn.net/?f=dx%20%3D%203%5Ccos%7B%5Ctheta%7Dd%5Ctheta)
So
![\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7B3%5Ccos%7B%5Ctheta%7Dd%5Ctheta%7D%7B%5Csqrt%7B9-%283%5Csin%7B%5Ctheta%7D%29%5E2%7D%7D%20%3D%20%5Cint%20%5Cfrac%7B3%5Ccos%7B%5Ctheta%7Dd%5Ctheta%7D%7B%5Csqrt%7B9%20-%209%5Csin%5E%7B2%7D%7B%5Ctheta%7D%7D%7D%20%3D%20%5Cint%20%5Cfrac%7B3%5Ccos%7B%5Ctheta%7Dd%5Ctheta%7D%7B%5Csqrt%7B9%281%20-%20%5Csin%5E%7B%5Ctheta%7D%29%7D%7D)
We have the following trigonometric identity:
![\sin^{2}{\theta} + \cos^{2}{\theta} = 1](https://tex.z-dn.net/?f=%5Csin%5E%7B2%7D%7B%5Ctheta%7D%20%2B%20%5Ccos%5E%7B2%7D%7B%5Ctheta%7D%20%3D%201)
So
![1 - \sin^{2}{\theta} = \cos^{2}{\theta}](https://tex.z-dn.net/?f=1%20-%20%5Csin%5E%7B2%7D%7B%5Ctheta%7D%20%3D%20%5Ccos%5E%7B2%7D%7B%5Ctheta%7D)
Replacing into the integral:
![\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7B3%5Ccos%7B%5Ctheta%7Dd%5Ctheta%7D%7B%5Csqrt%7B9%281%20-%20%5Csin%5E%7B2%7D%7B%5Ctheta%7D%29%7D%7D%20%3D%20%5Cint%7B%5Cfrac%7B3%5Ccos%7B%5Ctheta%7Dd%5Ctheta%7D%7B%5Csqrt%7B9%5Ccos%5E%7B2%7D%7B%5Ctheta%7D%7D%7D%20%3D%20%5Cint%20%5Cfrac%7B3%5Ccos%7B%5Ctheta%7Dd%5Ctheta%7D%7B3%5Ccos%7B%5Ctheta%7D%7D%20%3D%20%5Cint%20d%5Ctheta%20%3D%20%5Ctheta%20%2B%20C)
Coming back to x:
We have that:
![x = 3\sin{\theta}](https://tex.z-dn.net/?f=x%20%3D%203%5Csin%7B%5Ctheta%7D)
So
![\sin{\theta} = \frac{x}{3}](https://tex.z-dn.net/?f=%5Csin%7B%5Ctheta%7D%20%3D%20%5Cfrac%7Bx%7D%7B3%7D)
Applying the arcsine(inverse sine) function to both sides, we get that:
![\theta = \arcsin{(\frac{x}{3})}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Carcsin%7B%28%5Cfrac%7Bx%7D%7B3%7D%29%7D)
The result of the integral is:
![\arcsin{(\frac{x}{3})} + C](https://tex.z-dn.net/?f=%5Carcsin%7B%28%5Cfrac%7Bx%7D%7B3%7D%29%7D%20%2B%20C)