Question #e61aa

2 Answers
Dec 7, 2017

Q1: #=-sqrt(4-x^2)/x^2 - sin^-1(x/2) + C#

Explanation:

Integrals can be solved in many ways, but for these I'll be using trig-substitution for Q1 and partial fractions for Q2.

Q1: #intsqrt(4-x^2)/(x^2)dx#
For this, we want to make use of the fact that #1-sin^2x=cos^2x#, so that we can make that #sqrt# sign at the top go away.

So, instead of just using #x=sintheta#, we'll use #x=2sintheta#, and you'll see why when we sub it in:

Let #x=2sintheta#
#dx/(d theta)=2costheta#
#dx=2costheta*d theta#

#intsqrt(4-x^2)/(x^2)dx=intsqrt(4-4sin^2 theta)/(4sin^2theta) * 2costheta d theta#
#=2int(2sqrt(1-sin^2 theta))/(4sin^2theta) *costheta d theta#
#=int(sqrt(cos^2 theta))/(sin^2theta) *costheta d theta#
#=int(cos theta)/(sin^2theta) *costheta d theta#
#=int(cos^2 theta)/(sin^2theta) d theta#
#=int cot^2 theta d theta#
#=int csc^2 theta - 1 d theta#
#=-cot theta - theta + C#

Even though we've done the integration, we have to change it back into terms of #x#. To so this, we make an observation from our initial substitution:

#x=2sintheta#
#sintheta = x/2#, and from here we can draw a right angle triangle with hypotenuse #2# and side opposite to #theta# as #x# (Idk how to insert a diagram here soz)
This leaves us with #cot theta = sqrt(4-x^2)/x^2#, from the triangle.
Also, #theta = sin^-1(x/2)#

Therefore, subbing it into the achieved result we get:
#=-sqrt(4-x^2)/x^2 - sin^-1(x/2) + C#

Dec 7, 2017

Q2: #1/6ln|(x-3)/(x+3)|+C#
Q1 is below all of this

Explanation:

Q2: #int(1)/(x^2-9)dx#

First, we break up the fraction inside:
#(1)/(x^2-9) = (1)/((x+3)(x-3))#
#-=A/(x-3)+B/(x+3)#, where #A# and #B# are numbers which we have to find the value of.
To do this, we can multiply both sides by #(x-3)(x+3):#
#1-=A(x+3)+B(x-3)#
Then to solve for #A# and #B#, we can sub in #x=3# and #x=-3#:

Sub in #x=3#:
#1-=A(6)+B(0)#
#A=1/6#

Sub in #x=-3#:
#1-=A(0)+B(-6)#
#B=-1/6#

Therefore, we see that:
#(1)/(x^2-9) =(1/6)/(x-3)+(-1/6)/(x+3)#

We will use this in our integration:
#int(1)/(x^2-9)dx= int(1/6)/(x-3)+(-1/6)/(x+3)dx#
#=1/6int(1)/(x-3)-(1)/(x+3)dx#
Recognize that each of the fractions is a #ln# integral (if this needs further explaining plz comment so I can do so)
#=1/6(ln|x-3|-ln|x+3|)+C#
#=1/6ln|(x-3)/(x+3)|+C#