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$\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$

Effort;

$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$

$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$

$\displaystyle\int\frac{1-t^2}{t}\cdot\frac{2t}{\sqrt{t^4-2t^2}}dt=2\int\frac{1-t^2}{\sqrt{t^4-2t^2}}dt$

$\ = 2\displaystyle\int\frac{1}{\sqrt{t^4-2t^2}}dt-2\displaystyle\int\frac{t}{\sqrt{t^2-2t}}dt$

$\ = 2\displaystyle\int t^{-1}(t^2-2)^{-\frac{1}{2}}dt-2\displaystyle\int t(t^2-2t)^{-\frac{1}{2}}dt$

But after that I don't know how to continue.

J. M. ain't a mathematician
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Micheal Brain Hurts
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4 Answers4

4

By setting $x=\frac{\pi}{2}-t$ the problem boils down to finding: $$ \int \frac{\cos t}{\sqrt{1-\cos t}}\,dt = \frac{1}{\sqrt{2}}\int\frac{1-2\sin^2\frac{t}{2}}{\sin\frac{t}{2}}\,dt $$ where: $$ \int \frac{1}{\sin\frac{t}{2}}\,dt = C + 2\log\left(\tan\frac{t}{4}\right).$$

Jack D'Aurizio
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3

Slightly different attempt to get rid of the trig functions (assuming a domain where no sign issues cause trouble):

$$\begin{array}{rl} \displaystyle \frac{\sin x}{\sqrt{1-\sin x}} & \displaystyle = \frac{\sin x\sqrt{1+\sin x}}{\sqrt{1-\sin x}\sqrt{1+\sin x}}\\[7 pt] & \displaystyle = \frac{\sin x\sqrt{1+\sin x}}{\sqrt{1-\sin^2 x}}\\[7 pt] & \displaystyle = \frac{\cos x \sin x\sqrt{1+\sin x}}{\cos^2 x}\\[7 pt] & \displaystyle = \frac{\cos x \sin x\sqrt{1+\sin x}}{1-\sin^2 x} \end{array}$$

Now let $t = \sin x$:

$$\int \frac{\sin x}{\sqrt{1-\sin x}} \,\mbox{d}x =\int \frac{\cos x \sin x\sqrt{1+\sin x}}{1-\sin^2 x}\,\mbox{d}x \to \int \frac{t\sqrt{1+t}}{1-t^2} \,\mbox{d}t$$

This can be rationalized with $u^2 = 1+t$ to get (after simplifying):

$$\int \left( -2 - \frac{2}{u^2-2} \right) \, \mbox{d}u$$

This appears to be a longer route than the suggestions given in some other answers ;o).

StackTD
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1

HINT:

$$\dfrac{\sin x}{\sqrt{1-\sin x}}=-\sqrt{1-\sin x}+\dfrac1{\sqrt{1-\sin x}}$$

$$1-\sin x=1-\cos\left(\dfrac\pi2-x\right)=2\sin^2\left(\dfrac\pi4-\dfrac x2\right)$$

Now for real $a,$ $$\sqrt{a^2}=|a|=\begin{cases}+a &\mbox{if } a\ge0 \\-a & \mbox{if } a<0 \end{cases}$$

lab bhattacharjee
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I might be missing something, but I think this is possible to do with a prolonged substitution:

$$ \int \frac{\sin(x)}{\sqrt{1-\sin(x)}} \ dx. $$

Let $u = 1-\sin(x)$, then $$ du = -\cos(x) dx \implies -\sec(x) du = dx.$$

Then we get $$ \int (1-u) \frac{1}{\sqrt{u}} (-\sec(x) du).$$

We need to find $\sec(x)$ in terms of $u$:

$$ u = 1-\sin(x) \\ \sin(x) = 1-u \\ \sin^2(x) = (1-u)^2 \\ 1-\cos^2(x) = 1-2u+u^2 \\ \cos^2(x) = 2u-u^2 \\ \cos(x) = \sqrt{2u-u^2 }\\ \sec(x) = \frac{1}{\sqrt{2u-u^2}} $$

So the integral becomes $$ \int \frac{u-1}{\sqrt{u}} \cdot \frac{1}{\sqrt{2u-u^2}} \ du \\ =\int \frac{u-1}{\sqrt{u^2}} \cdot \frac{1}{\sqrt{2-u}} \ du \\ =\int \frac{u-1}{u} \cdot \frac{1}{\sqrt{2-u}} \ du \\ =\int \frac{1}{\sqrt{2-u}} \ du - \int \frac{1}{u\sqrt{2-u}} \ du $$

The first term is easy to evaluate: $\int \frac{1}{\sqrt{2-u}} \ du = -2\sqrt{2-u}$.

The second term is a bit more involved but with a substitution $v = \sqrt{2-u}$ you'll find that you can use inverse hyperbolic trig to get $\int \frac{1}{u\sqrt{2-u}} \ du = -\sqrt{2} \tanh^{-1}( \frac{\sqrt{2-u}}{\sqrt{2}})$. So altogether, we've got

$$ \int \frac{1}{\sqrt{2-u}} \ du - \int \frac{1}{u\sqrt{2-u}} \ du \\ = -2\sqrt{2-u} - \left( \sqrt{2} \tanh^{-1} \left( \frac{\sqrt{2-u}}{\sqrt{2}}\right) \right) + C \\ = \sqrt{2} \tanh^{-1} \left( \frac{\sqrt{2-u}}{\sqrt{2}} \right) - 2\sqrt{2-u} +C. $$

And now to replace $u$ with $1-\sin(x)$, $$\sqrt{2} \tanh^{-1} \left( \frac{\sqrt{2-u}}{\sqrt{2}} \right) - 2\sqrt{2-u} +C \\ =\sqrt{2} \tanh^{-1} \left( \frac{\sqrt{2-(1-\sin(x))}}{\sqrt{2}} \right) - 2\sqrt{2-(1-\sin(x))} +C \\ = \sqrt{2} \tanh^{-1} \left( \frac{\sqrt{\sin(x)+1}}{\sqrt{2}} \right) - 2\sqrt{\sin(x)+1} +C $$

Xoque55
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