Question

√1-sinx dx

Answer 1

Answer on Question #37728 - Math - Integral Calculus

Solution.

Take the integral:

$$\int \sqrt {1 - \sin x} \, dx$$

Substitution:

$$u = 1 - \sin x \quad \rightarrow \quad \sin x = 1 - u$$ $$d u = - \cos x \, dx \quad \rightarrow \quad d x = - \frac {d u}{\cos x}$$ $$\cos x = \sqrt {1 - (\sin x) ^ {2}} = \sqrt {1 - (1 - u) ^ {2}} = \sqrt {1 - 1 + 2 u - u ^ {2}} = \sqrt {u (2 - u)}$$

that's mean

$$d x = - \frac {d u}{\cos x} = - \frac {d u}{\sqrt {u (2 - u)}}$$

So we can rewrite the integral

$$\int \sqrt {1 - \sin x} \, dx = - \int \sqrt {u} \cdot \frac {d u}{\sqrt {u (2 - u)}} = - \int \frac {d u}{\sqrt {2 - u}} =$$

Substitute:

$$s = 2 - u$$ $$d s = - d u$$ $$= \int \frac {1}{\sqrt {s}} \, ds = 2 \sqrt {s} + \text{constant} =$$

Substitute back for $s = 2 - u$

$$= 2 \sqrt {2 - u} + \text{constant} =$$

Substitute back for $u = 1 - \sin x$

$$= 2 \sqrt {2 - 1 + \sin x} + \text{constant} = 2 \sqrt {1 + \sin x} + C$$

Answer:

$$\int \sqrt {1 - \sin x} \, dx = 2 \sqrt {1 + \sin x} + C$$