Intergration (arcsinx)^2 ?

1 Answer
May 24, 2018

# x(arcsinx)^2+2(sqrt(1-x^2))arcsinx-2x+C#.

Explanation:

Let, #I=int(arcsinx)^2dx#.

Subst. #arcsinx=t. :. x=sint. :. dx=costdt#.

Hence, #I=intt^2costdt#.

We use the following Rule of Integration by Parts (IBP) :

#intuv'dt=uv-intu'vdt," where, "u=t^2, v'=cost#, so that,

# u'=2t, v=intv'dt=intcostdt=sint#.

#:. I=t^2sint-int2t*sintdt=t^2sint-2I_1#, where,

#I_1=inttsintdt#.

We again use IBP with #u=t, v'=sint; :. u'=1, v=-cost#,

#:. I_1=t(-cost)-int{(1)(-cost)}dt#,

#=-tcost+intcostdt=-tcost+sint#.

Utilising #I_1# in #I#, we get,

#I=t^2sint-2{-tcost+sint}#,

#=t^2sint+2tcost-2sint#,

#=t^2sint+2tsqrt(1-sin^2t)-2sint#.

Reverting from #t" to "x," by, "t=arcsinx, &, sint=x#,

#I=x(arcsinx)^2+2(sqrt(1-x^2))arcsinx-2x+C#.

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