bemath-sin-x-1-sin-x-dx-

Question Number 112271 by bemath last updated on 07/Sep/20

\sqrt{bemath}

\int \sin x \sqrt{1 - \sin x} dx ?

Answered by maths mind last updated on 07/Sep/20

1 - s i n (x) = {(s i n (\frac{x}{2}) - c o s (\frac{x}{2}))}^{2}

Commented by bemath last updated on 07/Sep/20

\sqrt{1 - \sin x} = ∣ \sin \frac{x}{2} - \cos \frac{x}{2} ∣

Answered by MJS_new last updated on 07/Sep/20

\int \sin x \sqrt{1 - \sin x} d x =

[t = \sin x \to d x = \frac{d t}{\cos x}]

= \int \frac{t \sqrt{1 - t}}{\sqrt{1 - t^{2}}} d t = \int \frac{t}{\sqrt{1 + t}} d t = \frac{2}{3} (t - 2) \sqrt{t + 1}

but {we}^{'} re losing sign changes \dots

other idea

\int \sin x \sqrt{1 - \sin x} d x =

[t = \frac{x}{2} - \frac{π}{4} \to d x = 2 d t]

= 2 \int \cos 2 t \sqrt{1 - \cos 2 t} d t =

= 2 \int ({2 \cos}^{2} t - 1) \sqrt{1 - ({2 \cos}^{2} t - 1)} d t =

= 2 \sqrt{2} \int ({2 \cos}^{2} t - 1) \sqrt{1 - \cos^{2} t} d t =

= 2 \sqrt{2} \int ({2 \cos}^{2} t \sin t - \sin t) d t =

= \sqrt{2} \int (\sin 3 t - \sin t) d t =

= \sqrt{2} \cos t - \frac{\sqrt{2}}{3} \cos 3 t = \dots

but {we}^{'} re also losing sign changes

Commented by MJS_new last updated on 07/Sep/20

found it! we must somehow keep the factor

\sqrt{1 - \sin x} in the result .

\int \sin x \sqrt{1 - \sin x} d x =

[t = \sin x \to d x = \frac{d t}{\cos x}]

= \int \frac{t \sqrt{1 - t}}{\sqrt{1 - t^{2}}} d t =

[u = \frac{\sqrt{1 - t}}{\sqrt{1 - t^{2}}} \to d t = - \frac{2 (t + 1) \sqrt{1 - t^{2}}}{\sqrt{1 - t}}]

= 2 \int (\frac{1}{u^{2}} - \frac{1}{u^{4}}) d u = \frac{2}{3 u^{3}} - \frac{2}{u} =

= \frac{2 (t - 2) \sqrt{1 - t^{2}}}{3 \sqrt{1 - t}} =

= \frac{2 (\sin x - 2) \cos x}{3 \sqrt{1 - \sin x}} + C

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