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Evaluate $$I=\int \frac{\sec x \:dx}{\sin (2x+\theta)+\sin \theta}$$

I used $$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$$ we get

$$I=\frac{1}{2} \times \int \frac{\sec x \:dx}{\sin (x+\theta) \cos x}=\frac{1}{2} \times \int \frac{\sec^2 x \:dx}{\sin (x+\theta)} $$

Now i used $\tan x=y$ we get

$$2I=\int \frac{\sqrt{1+y^2}dy}{y \cos \theta+\sin \theta}$$

But how can we proceed from here?

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  • $\begingroup$ You can Write $y\cos \theta+\sin \theta = \sqrt{1+y^2}\left[\cos \left(\theta-\alpha \right)\right]$ $\endgroup$ – juantheron May 11 '16 at 7:09
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    $\begingroup$ but $\alpha$ is a function of $y$ again right? $\endgroup$ – Umesh shankar May 11 '16 at 7:11
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    $\begingroup$ Wolfram gives a really complicated answer, look at wolframalpha.com/input/… $\endgroup$ – jim May 11 '16 at 7:36

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