How do I integrate:

$$\int \frac{\sec x}{\sqrt{\sin(2x + A) + \sin A}}\, dx?$$ First, I tried to substitute $t^2$ for the denominator, but it was really a great flop. I then removed $\sin A$ since it is a constant but then the integrand was not in good shape. Can anyone help me? Any hint(s) will be appreciated. Don't solve it; it is my job! Just some help.


Let $$\displaystyle I = \int\frac{\sec x}{\sqrt{\sin 2x\cos A+\cos 2x\cdot \sin A+\sin A}}dx$$

$$\displaystyle I = \int\frac{\sec x}{\sqrt{2\sin x\cos x \cdot \cos A+\sin A\cdot 2\cos^2 x}}dx$$

So $$\displaystyle I =\frac{1}{\sqrt{2}}\int\frac{\sec x}{\cos x\sqrt{\cos A\cdot \tan x+\sin A}}dx$$

$$\displaystyle I = \frac{1}{\sqrt{2}}\int\frac{\sec ^2x}{\sqrt{\cos A\cdot \tan x+\sin A}}dx$$

Now Let $$\tan x\cdot \cos A+\sin A = t^2\;,$$ Then $$\displaystyle \sec^2x dx = \frac{2t}{\cos A}dt$$

So $$\displaystyle I = \frac{1}{\sqrt{2}\cos A}\int\frac{2t}{t}dt = \frac{\sqrt{2}}{\cos A}t+\mathcal{C} = \frac{\sqrt{2}}{\cos A}\left[\sqrt{\tan x\cdot \cos A+\sin A}\right]+\mathcal{C}$$

  • $\begingroup$ This is really, really clever! +$1$ $\endgroup$ – Clayton Aug 22 '15 at 17:35
  • $\begingroup$ Thanks! So, what should be my approach to solve these integrals? Just find a way to get tangent & then $\sec^2 x$, right?? BTW, your first step still is lamenting for $dx$ :p $\endgroup$ – user142971 Aug 22 '15 at 17:41
  • $\begingroup$ Yes You are right, Thanks for pointing error, $\endgroup$ – juantheron Aug 22 '15 at 17:42

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