Question

Question #e61aa

Answer 1

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`

Answer 2

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`