Question

`\int(x^2dx)/(\sqrt(4-9x^2))`?

I know it involves trigonometric substitution, and that
`\sqrt(a^2-x^2)` form with `x=a\sin\theta` (`-\pi/2\le\theta\le\pi/2`, `1-\sin^2x=\cos^2x`)

However, I am stuck at the substitution part here.
I know that `x=2/3\sin\theta`, `\thereforea=2/3`, but not what to do afterwards.

Would it be `dx=2/3\cos\theta`?

Answer 1

`int(x^2dx)/sqrt(4-9x^2)=2/27(arcsin(3/2x)-(3xsqrt(4-9x^2))/4)+C`

Explanation:

For an integral involving the root `sqrt(a^2-b^2x^2),` we may use the substitution

`x=a/bsintheta`

So, here, `a=2, b=3, x=2/3sintheta, dx=2/3costhetad theta`

Thus, we have

`(2/3)^2(2/3)int(sin^2thetacostheta)/sqrt(4-4sin^2theta)d theta`

`(4/9)(cancel2/3)(1/cancel2)int(sin^2thetacostheta)/sqrt(1-sin^2theta)d theta`

`1-sin^2theta=cos^2theta,` so

`4/27int(sin^2thetacostheta)/sqrt(cos^2theta)d theta`

`4/27int(sin^2thetacancel(costheta))/cancel(costheta)d theta`

Recalling that `sin^2theta=1/2(1-cos2theta)`,

`4/27intsin^2thetad theta=4/27(1/2)int(1-cos2theta)d theta`

`2/27(theta-1/2sin2theta)+C`

We need things in terms of `x.`

Since `x=2/3sintheta, sintheta=3/2x, theta=arcsin(3/2x)`

Furthermore, recall that `1/2sin2theta=sinthetacostheta`. We'll need to find the cosine using a quick sketch. Since `sintheta=(3x)/2,` we label the opposite and hypotenuse as follows, realizing that the adjacent side will be the radical originally found in the integral:

We see `costheta=sqrt(4-9x^2)/2, 1/2sin2theta=(3/2x)sqrt(4-9x^2)/2=(3xsqrt(4-9x^2))/4`

Thus,

`int(x^2dx)/sqrt(4-9x^2)=2/27(arcsin(3/2x)-(3xsqrt(4-9x^2))/4)+C`

Answer 2

`"Sometimes it is better to avoid"color(red)" trigonometric substitution"` `"and think about"color(blue)" another substitution "`that led us to known formula.
`(1)int1/sqrt(a^2-x^2)dx=sin^-1(x/a)+c`

Explanation:

#color(red)((2)intsqrt(a^2-x^2)dx=x/2sqrt(a^2- x^2)+a^2/2sin^-1(x/a)+c#

Here,

`I=intx^2/sqrt(4-9x^2)dx`

Let, `3x=t=>3dx=dt=>dx=1/3dt,and9x^2=t^2=>x^2=t^2/9`

`:.I=int((t^2/9))/sqrt(4-t^2)1/3dt`

`=1/27intt^2/sqrt(4-t^2)dt`

`=-1/27[int(-t^2)/sqrt(4-t^2)dt]`

`=-1/27[int(4-t^2-4)/sqrt(4-t^2)dt]`

`=-1/27[int(4-t^2)/sqrt(4-t^2)dt-int4/sqrt(4-t^2)dt]`

`=-1/27[intsqrt(4-t^2)dt-4int1/sqrt(4-t^2)dt]`

#=-1/27[intsqrt(2^2-t^2)dt-4int1/sqrt(2^2-t^2)dt]toApply(1)& (2)#

`=-1/27[t/2sqrt(2^2-t^2)+2^2/2sin^-1(t/2)-4sin^-1(t/2)]+c`

`=-1/27[t/2sqrt(4-t^2)+2sin^-1(t/2)-4sin^-1(t/2)]+c`

`=-1/27[t/2sqrt(4-t^2)-2sin^-1(t/2)]+c`

`=1/27[2sin^-1(t/2)-t/2sqrt(4-t^2)]+c`

Subst. back, `..tot=3x,t^2=9x^2`

`=1/27[2sin^-1((3x)/2)-(3x)/2sqrt(4-9x^2)]+c`