sin1(y)dy\int \sin^{-1}{(y)} \, dy

1
Use Integration by Parts on sin1(y)dy\int \sin^{-1}{(y)} \, dy.
Let u=sin1(y)u=\sin^{-1}{(y)}, dv=1dv=1, du=11y2dydu=\frac{1}{\sqrt{1-{y}^{2}}} \, dy, v=yv=y

2
Substitute the above into uvvduuv-\int v \, du.
(sin1(y))yy1y2dy(\sin^{-1}{(y)})y-\int \frac{y}{\sqrt{1-{y}^{2}}} \, dy

3
Use Integration by Substitution on y1y2dy\int \frac{y}{\sqrt{1-{y}^{2}}} \, dy.
Let u=1y2u=1-{y}^{2}, du=2ydydu=-2y \, dy, then ydy=12duy \, dy=-\frac{1}{2} \, du

4
Using uu and dudu above, rewrite y1y2dy\int \frac{y}{\sqrt{1-{y}^{2}}} \, dy.
12udu\int -\frac{1}{2\sqrt{u}} \, du

5
Use Constant Factor Rule: cf(x)dx=cf(x)dx\int cf(x) \, dx=c\int f(x) \, dx.
121udu-\frac{1}{2}\int \frac{1}{\sqrt{u}} \, du

6
Since 1x=x12\frac{1}{\sqrt{x}}={x}^{-\frac{1}{2}}, using the Power Rule, x12dx=2x12\int {x}^{-\frac{1}{2}} \, dx=2{x}^{\frac{1}{2}}
u-\sqrt{u}

7
Substitute u=1y2u=1-{y}^{2} back into the original integral.
1y2-\sqrt{1-{y}^{2}}

8
Rewrite the integral with the completed substitution.
(sin1(y))y+1y2(\sin^{-1}{(y)})y+\sqrt{1-{y}^{2}}

9
Add constant.
(sin1(y))y+1y2+C(\sin^{-1}{(y)})y+\sqrt{1-{y}^{2}}+C

Done

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