Question

Solve following integral?

a.) `∫sin^-1(2x)dx`

Answer 1

The answer is `=xarcsin(2x)+1/2sqrt(1-4x^2)+C`

Explanation:

We need

`int(u'(x)dx)/(sqrt(u(x)))=2sqrt(u(x))+C`

We solve this integral by integration by parts.4

`intuv'dx=uv-intu'v`

Here,

`u=arcsin(2x)`, `=>`, `u'=2/sqrt(1-(2x)^2)`

`v'=1`, `=>`, `v=x`

Therefore,

`intarcsin(2x)dx=xarcsin(2x)-int(2xdx)/(sqrt(1-4x^2))`

`=xarcsin(2x)+1/4*2sqrt(1-4x^2)+C`

`=xarcsin(2x)+1/2sqrt(1-4x^2)+C`

Answer 2

The answer `=1/2*sin^-1(2x)` orr `=-1/2*cos^-1(2x)+c`

Explanation:

Note that

`sin^-1(x)=1/sqrt(1-x^2)`

now let solve

`∫sin^-1(2x)dx=int[1/sqrt(1-(2x)^2]]*dx`

`=-1/2*cos^-1(2x)+c`

orrrrrrrrrr

`=1/2*sin^-1(2x)`

Answer 3

`=>x arcsin(2x)+sqrt(1-4x^2)/(2)+C`

Explanation:

Given,
`int sin^-1 (2x) dx`
Substitute,
`color(grey)(u=2x->dx=(1/2)du`
`=(1/2)int sin^-1(u)du`
Now,
`=int sin^-1(u)du`
Now,
Integrate by parts:
`color(grey)(=>intfg'=fg-intf'g)`
`color(grey)(where, f=>sin^-1(u)*g'=1`
`color(grey)(and, f'=>(1)/(sqrt(1-u^2))*g=v`
Therefore,
`=>usin^-1(u)-intu/(sqrt(1-u^2)) du`

Now,
`intu/(sqrt(1-u^2))du`
Substitute,
`color(grey)(v=1-u^2->du=-(1/(2v))dv`
Now,
`int1/(sqrtv) dv`
Apply power rule,
`color(grey)(intv^n dv=(v^n+1)/(n+1), with, n=-1/2`
Therefore, we get,
`=2sqrtv`

Putting in the solved integrals,
`=>-1/2int(1)/(sqrt v)dv`
`=-sqrtv`
Now,
Undo substitution,
`color(grey)(v=1-u^2)`
Putting in solved integrals,
`=>u sin^-1(u)-int(u)/(sqrt(1-u^2))(du)`
`=>u sin^-1(u)+sqrt(1-u^2)`
Now,
Putting in solved integrals,
`=>1/2int sin^-1(u) du`
`=>u sin^-1(u)/(2)+sqrt(1-u^2)/(2)`
Now,
Undo substitution,
`=>color(grey)(u=2x)`
`therefore x sin^-1(2x)+sqrt(1-4x^2)/(2)+C`
`Phew...`