3 answers

Solve following integral?

Question:

a.) #∫sin^-1(2x)dx#


Answers

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#

.

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)#

.

#=>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...#

.

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