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I need to integrate the following function:$$\frac{\cos(x)}{1+\sin(x)}\sqrt{\frac{2-\sin(x)}{2+\sin(x)}}$$

For this I use the substitution $2+\sin(x)=y^2$

I get $$2\int{\frac{\sqrt{4-y^2}}{y^2-1}} dy$$.

Then I use $y=2\sin(z)$.

So I get:

$$I=8\int\frac{{\cos^2(z)}}{4\sin^2(z)-1}dz$$

$$\implies I=2\left[\frac{\cos^2(z)}{\sin^2(z)-\frac{1}{4}}\right]$$

$$-2z+\frac{3}{2}\int{\frac{1}{\sin^2(z)-\frac{1}{4}}}dz + C $$

After this, how to integrate $$\int{\frac{1}{\sin^2(z)-\frac{1}{4}}}dz$$ (without using any hypergeometric functions) ? You may use logarithmic functions.

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    Do you mean to say that you want to integrate this without using trigonometric identities? – 2017-01-05
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    the Weierstrass substitution will save your day (why even think about something as complicated as hypergeometric functions?) – 2017-01-05
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    @tired Yes, that works. Thanks. (Well, that happens when one revisits Integral Calculus after a very long time :P) – 2017-01-05

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Well, multiply numerator and denominator by $4\csc^2(z)$:

$$\int\frac{1}{\sin^2(z)+\frac{1}{4}}\space\text{d}z=\int\frac{4\csc^2(z)}{4\csc^2(z)\left(\sin^2(z)+\frac{1}{4}\right)}\space\text{d}z=\int\frac{4\csc^2(z)}{\cot^2(z)+5}\space\text{d}z$$

Now, substitute $u=\cot(z)$:

$$\int\frac{4\csc^2(z)}{\cot^2(z)+5}\space\text{d}z=-4\int\frac{1}{u^2+5}\space\text{d}u=\text{C}-\frac{4\arctan\left(\frac{u}{\sqrt{5}}\right)}{\sqrt{5}}$$

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    Isn't it supposed to be $-\frac 14$ in the denominator? – 2017-01-05
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    @DavidQuinn I agree. Though it will just affect some signs but I guess the method will still work. – 2017-01-05
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    Why would you want to accept an answer which is actually wrong? – 2017-01-05
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    I will re-accept this answer after it is edited (which @DavidQuinn pointed out). – 2017-01-05