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$$\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}$$

$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx$$

I am stuck. Please help me....

Empty
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Brahmagupta
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  • Please improve the question by providing additional context, which ideally includes motivation for the integral, your thoughts on the problem, and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Carl Mummert Aug 13 '15 at 23:16
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    Seriously asked, seriously answered. Up-voted. Why the 'close' votes. – BruceET Aug 19 '15 at 03:25

2 Answers2

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Hint:

$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$$

$$=\int\frac{\cos x\sqrt{4-\sin^2 x}}{(1+\sin x)(2+\sin x)}\,dx$$

( multiplying numerator & denominator by $(1+\sin x)(2+\sin x)$ under square root sign.)

Now put , $\sin x=z$.

Expand Hint :

Then ,

$$=\int\frac{1}{1+z}\sqrt{\frac{2-z}{2+z}}\,dz$$

$$=\int u\sqrt{\frac{3u-1}{u+1}}\,du\text{ , by putting $1+z=\frac{1}{u}.$ }$$

$$=\int\frac{u(3u-1)}{\sqrt{(u+1)(3u-1)}}\,du$$

$$=\int\sqrt{3u^2+2u-1}\,du-\frac{1}{2}\int \frac{d(3u^2+2u-1)}{\sqrt{3u^2+2u-1}}+2\int\frac{\,du}{\sqrt{3u^2+2u-1}}$$

$$=\cdots \cdots \cdots \cdots \cdots $$

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Empty
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    I think I'm missing something obvious after that. How would I integrate the resulting function? – Gummy bears Aug 13 '15 at 16:52
  • @Gummy bears) Please see my updated answer........I think from here you can proceed... – Empty Aug 14 '15 at 02:34
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    Even though I +1'd this answer, there is an issue here I'm not very satisfied about: $\sqrt{1-\sin^2x} = |\cos x|$. This is hardly a problem when calculating definite integral as one can easily split domain of integration in parts where $\cos x$ is positive or negative, but this is really a question on indefinite integral. EDIT: perhaps one could make another trigonometric substitution to cancel out absolute value, something akin of $|\cos x||\cos x| = \cos^2 x$ and use a formula for $\cos\arcsin x$ or something similar. – Ennar Aug 14 '15 at 13:21
  • @@ Ennar ) For indefinite integral there is no need to use absolute value. – Empty Aug 14 '15 at 13:27
  • @Ennar's comment is actually quite relevant. Clearly the original integrand is always positive. But your line with $\cos x$ in the numerator has an integrand that is negative just as often as it is positive. The original expression's antiderivative will grow unbounded, but the expression with $\cos x$ in the numerator will lead to an antiderivative that is periodic. – 2'5 9'2 Aug 24 '15 at 16:53
6

Let $$\displaystyle I = \int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$$

We can write $$\displaystyle \sqrt{\frac{1-\sin x}{1+\sin x}} = \sqrt{\frac{1-\sin x}{1+\sin x}\times \frac{1+\sin x}{1+\sin x}} = \frac{\cos x}{1+\sin x}$$

So we get $$\displaystyle I = \int\frac{\cos x}{1+\sin x}\cdot \sqrt{\frac{2-\sin x}{2+\sin x}}dx$$

Now Let $1+\sin x= y\;,$ Then $\cos xdx = dy$

So Integral $$\displaystyle I = \int\frac{1}{y}\cdot \sqrt{\frac{3-y}{1+y}}dy$$

Now Put $$\displaystyle \frac{3-y}{1+y}=t^2\Rightarrow y=\frac{3-t^2}{1+t^2}$$

So we get $$\displaystyle y=-\left[1-\frac{4}{1+t^2}\right] = \left[\frac{4}{1+t^2}-1\right].$$ So $\displaystyle dy = -\frac{8t}{(1+t^2)^2}$

So Integral $$\displaystyle I = \int\frac{1+t^2}{3-t^2}\cdot t\cdot \frac{-8t}{(1+t^2)^2}dt = 8\int\frac{t^2}{(t^2-3)\cdot (1+t^2)}dt$$

So Integral $$\displaystyle I = 2\int \left[\frac{3(t^2+1)+(t^2-3)}{(t^2-3)\cdot (1+t^2)}\right]dt = 2\int \left[\frac{3}{t^2-(\sqrt{3})^2}+\frac{1}{1+t^2}\right]dt$$

So Integral $$\displaystyle I = 6\cdot \frac{1}{2\sqrt{3}}\cdot \ln\left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+2\tan^{-1}(t)+\mathcal{C}$$

So Integral $$\displaystyle I = \sqrt{3}\cdot \ln\left|\frac{\sqrt{2-\sin x}-\sqrt{3}\cdot \sqrt{2+\sin x}}{\sqrt{2-\sin x}-\sqrt{3}\cdot \sqrt{2+\sin x}}\right|+2\tan^{-1}\left(\sqrt{\frac{2-\sin x}{2+\sin x}}\right)+\mathcal{C}$$

juantheron
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