# Evaluate the integration : $\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$ [closed]

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

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

• 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. Aug 13, 2015 at 23:16

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 ,

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

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

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

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

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

• I think I'm missing something obvious after that. How would I integrate the resulting function? Aug 13, 2015 at 16:52
• @Gummy bears) Please see my updated answer........I think from here you can proceed... Aug 14, 2015 at 2:34
• 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. Aug 14, 2015 at 13:21
• @@ Ennar ) For indefinite integral there is no need to use absolute value. Aug 14, 2015 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. Aug 24, 2015 at 16:53

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