Question

How do I evaluate `int (arcsin x)/sqrt (1+x) dx`?

Answer 1

We will integrate this using integration by parts.

Recall that `uv-intvdu`

Let `u=arcsin(x)` and `dv=dx/sqrt(1+x)`

Differentiating `u=arcsinx` we have
`du=1/sqrt(1-x^2)dx`

Integrate `dv=dx/sqrt(1+x)`

`intdv=intdx/sqrt(1+x)`

`v=2sqrt(1+x)`

Therefore,

`arcsin(x)(2sqrt(1+x))-int2sqrt(1+x)(1/sqrt(1-x^2))dx`

Some rewriting and factor `1-x^2`

`2arcsin(x)sqrt(1+x)- 2intsqrt(1+x)/sqrt((1+x)(1-x))dx`

`2arcsin(x)sqrt(1+x)-2intsqrt(1+x)/(sqrt(1+x)sqrt(1-x))dx`

`2arcsin(x)sqrt(1+x)-2int1/sqrt(1-x)dx`

Finally, we integrate

`2arcsin(x)sqrt(1+x)+4sqrt(1-x)+C`