2017年1月23日 · x^2+7x-18 Each term in the second bracket must be multiplied by each term in the first bracket. This can be achieved as follows. rArr(color(red)(x+9))(x-2) =color(red)(x)(x-2)color(red)(+9)(x-2) distribute the brackets. =x^2-2x+9x-18 …

2015年8月21日 · In other words, in oerder for the function to be defined, you need the expression that's under the square root to be positive. 9 - x^2 >= 0 x^2 <= 9 implies |x| <= 3 This means that you have x >= -3" " and " "x<=3 For any value of x outside the interval [-3, 3], the expression under the square root will be negative, which means that the ...

2018年3月20日 · (x-3)(x+3) >x^2-9" is a "color(blue)"difference of squares" "and in general factorises as" •color(white)(x)a^2-b^2=(a-b)(a+b) "here "a=x" and "b=3 rArrx^2-9=(x-3)(x+3)

2015年6月6日 · I got 9/2arcsin(x/3) - x/2sqrt(9-x^2) + C. You can do this with trig substitution. Notice how this is of the form sqrt(a^2-x^2), which looks like sqrt(1 - sin^2theta), while sin^2theta + cos^2theta = 1. So, let: a = sqrt9 = 3 x = asintheta = 3sintheta dx = acosthetad theta = 3costhetad theta That gives sqrt(9 - x^2) = sqrt(3^2 - (3sintheta)^2) = sqrt9sqrt(1-sin^2theta) = …

2016年2月28日 · Because #f(x,y)=sqrt(9-x^2-y^2)# we must have that #9-x^2-y^2>=0=>9>=x^2+y^2=>3^2>=x^2+y^2# The domain of #f(x,y)# is the border and the interior of the circle #x^2+y^2=3^2# or. The domain is represented by the disc whose center is the origin of the coordinates system and the radius is 3. Now hence #f(x,y)>=0# and #f(x,y)<=3# we find …

2016年12月3日 · The integration by parts formula is: int u(x)v(x) dx = u(x)v(x)-int v(x){du}/{dx}dx (or int udv = uv-int vdu if your prefer) so that v(x)=sin^{-1}(x/3) giving the integral shown in the answer. Notes: I assume that either you can quote the standard integral int 1/sqrt(a^2-x^2)dx=sin^{-1} (x/a) (or cos^{-1}(x/a), it makes little difference).

2018年3月30日 · In general, when encountering an integral involving sqrt(a^2-x^2), make the trigonometric substitution x=asintheta. In this case, a^2=9, a=3, and so our substitution is x=3sintheta. Solving for dx yields dx=3costhetad theta.

2018年4月1日 · Recalling that sin2theta=2sinthetacostheta, sin^2theta+cos^2theta=1: x^2/9+cos^2theta=9/9 cos^2theta=(9-x ...

How do you express # x^2/(x^2+9)^2# in partial fractions? Precalculus Matrix Row Operations Partial Fraction Decomposition (Linear Denominators) 1 Answer

2016年12月30日 · = 1/2 x sqrt (4- 9 x^2) + 2/3 sin^(-1) (3/2 x) + C One obvious thing here is to use a sub to get the integrand looking like this sqrt(4-4 sin^2 y) = 2 cos y, ie using the Pythagorean identity so we can say that 9 x^2 = 4 sin^2 y, and the sub we are going to try is x = 2/3 sin y, \\ dx = 2/3 cos y \\ dy The integration is then int 2 cos y * 2/3 cos y \\ dy = 4/3int …