Evaluate $ \int \frac{\sin^{-1}(x)}{\sqrt{1+x}} \; dx$

Question

So I have to find the integral of $$ \int \frac{\sin^{-1}(x)}{\sqrt{1+x}} \; dx$$

I think I have to do this using the integration by parts..so I will take $f = \sin^{-1}(x)$ and $ \sqrt {1+x}=g' $...what about now?

Answer 1

Do integration by parts with $f(x)=\arcsin(x)$ and $g(x)=\frac{1}{\sqrt{1+x}}$. Then $$ I=\int\frac{\arcsin x}{\sqrt{1+x}}\,\mathrm dx=\int f(x)g(x)\,\mathrm dx=f(x)G(x)-\int f'(x)G(x)\,\mathrm dx\\ =2\arcsin (x)\sqrt{1+x}-\int \frac{1}{\sqrt{1-x^2}}\cdot2\sqrt{1+x}\,\mathrm dx, $$ but $$ \frac{1}{\sqrt{1-x^2}}\cdot2\sqrt{1+x}=\frac{2\sqrt{1+x}}{\sqrt{1+x}\sqrt{1-x}}=\frac{2}{\sqrt{1-x}}, $$ so $$ I=2\arcsin (x)\sqrt{1+x}-\int \frac{2}{\sqrt{1-x}}\,\mathrm dx=2\arcsin (x)\sqrt{1+x}+4\sqrt{1-x}. $$

Answer 2

We can use the formula $$\int f g' dx = f \int g' dx - \int \frac{df}{dx} \left( \int g' dx\right)dx$$ Let $g' = \frac{1}{\sqrt{1+x}}$ and $f = \sin^{-1}(x)$ $$ \implies \int \frac{\sin^{-1}(x)}{\sqrt{1+x}} \; dx = \sin^{-1}(x) \int \frac{1}{\sqrt{1 + x^2}}dx - \int \frac{d}{dx} (\sin^{-1}(x)) \left( \int \frac{1}{\sqrt{1 + x^2}} dx \right) dx $$

Answer 3

Integrate by parts, letting $dv = (x+1)^{-1/2}$ and $u = \arcsin(x)$, then $du = \frac{1}{\sqrt{1-x^2}}$ and $v = \int (x+1)^{-1/2} dx = 2 \sqrt{x+1}$.

Use $\int u dv = uv - \int v du$ to reduce this to

$$ \int v du = \int 2 \sqrt{ \frac{x+1}{1-x^2} } dx = 2 \int \frac{dx}{\sqrt{1-x}} = -4 \sqrt{1-x}. $$