2016年11月5日 · #y=sqrt(9-x^2)# is the upper half of the circle #x^2+y^2=9#, a circle with radius #3# and center at #(0,0)#. Since the bounds span the length of the semicircle, the integral will …

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)

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 …

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 …

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 …

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 …

2016年7月13日 · x=1 or x=-5 (x+2)^2=9 can be written as (x+2)^2-9=0 or (x+2)^2-3^2=0 Now using the identity a^2-b^2=(a-b)(a+b), we have (x+2-3)(x+2+3)=0 or (x-1)(x+5)=0 Hence either …

2018年7月10日 · Let #x=3\sin\theta\implies dx=3\cos\theta\ d\theta# #\therefore \int (9-x^2)^{1/2}\ dx# #= \int (9-(3\sin\theta)^2)^{1/2}\ (3\cos\theta\ d\theta)#

2016年11月5日 · J=-intsqrt(9-x^2)dx Let x=3sintheta so that dx=3costhetad theta: J=-intsqrt(9-9sin^2theta)(3costhetad theta) J=-3intsqrt9sqrt(1-sin^2theta)(costheta)d theta J= …

2017年8月12日 · Domain: [-3,3] Range: [0,3] The value under a square root cannot be negative, or else the solution is imaginary. So, we need 9-x^2\\geq0, or 9\\geqx^2, so x\\leq3 and x\\geq …