Question

what is the integration of sin^-1 (x/(a+x))^0.5

Answer 1

Answer on Question #50525– Math – Integral Calculus

$$\int \sin^{-1} \sqrt{\frac{x}{a + x}} \, dx$$

Solution:

Integrating by parts

$$u = \sin^{-1} \sqrt{\frac{x}{a + x}}, \qquad du = \frac{1}{\sqrt{1 - \frac{x}{a + x}}} \cdot \frac{1}{2} \sqrt{\frac{a + x}{x}} \cdot \frac{a + x - x}{(a + x)^2} \, dx = \frac{\sqrt{a}}{2\sqrt{x}(a + x)} \, dx$$ $$dv = dx, \, v = x$$ $$\int \sin^{-1} \sqrt{\frac{x}{a + x}} \, dx = uv - \int v \, du = x \sin^{-1} \sqrt{\frac{x}{a + x}} - \int \frac{\sqrt{a}x}{2\sqrt{x}(a + x)} \, dx$$

Let $\sqrt{x} = t$, then $x = t^2$, and $\frac{dx}{2\sqrt{x}} = dt$, so $dx = 2\sqrt{x} \, dt = 2t \, dt$

Hence:

$$\begin{aligned} & \int \frac{\sqrt{a}x}{2\sqrt{x}(a + x)} \, dx = \int \frac{\sqrt{a}t}{2(a + t^2)} \cdot 2t \, dt = \sqrt{a} \int \frac{t^2 + a - a}{t^2 + a} \, dt = \\ & = \sqrt{a} \int dt - a\sqrt{a} \int \frac{dt}{t^2 + a} = \sqrt{a}t - \frac{a\sqrt{a}}{\sqrt{a}} \tan^{-1}\frac{t}{\sqrt{a}} + C = \\ & = \sqrt{a x} - a \tan^{-1}\frac{\sqrt{x}}{\sqrt{a}} + C, \end{aligned}$$

where $C$ is an arbitrary real constant.

Finally obtain:

$$\int \sin^{-1} \sqrt{\frac{x}{a + x}} \, dx = x \sin^{-1} \sqrt{\frac{x}{a + x}} - \sqrt{a x} + a \tan^{-1}\frac{\sqrt{x}}{\sqrt{a}} + C$$

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