Solve following integral?

a.) ∫sin^-1(2x)dxsin1(2x)dx

3 Answers
Apr 30, 2018

The answer is =xarcsin(2x)+1/2sqrt(1-4x^2)+C=xarcsin(2x)+1214x2+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

Apr 30, 2018

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)

Apr 30, 2018

=>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...