# The indefinite integral $\int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$

$$\int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$$

$$\int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}$$ $$\int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\frac{\sin x + \cos x}{\sqrt{\cos 2x}}}$$ $$\sqrt 2 \int \frac{\sin \left(2x +\alpha+ \frac{\pi}{4}\right) + \sin \left(\alpha -\frac{\pi}{4}\right)}{\cos^3 x\cdot\sqrt{\cos 2x}}$$

How I can do it after this and get rid of square root?

• I don't know if it helps, but I'd say that, since $\sin(x+a)=\sin x\cos\alpha+\cos x\sin\alpha$, a formula in $\alpha$ for those primitives is tantamount to the two cases $\alpha=0$ and $\alpha=\frac\pi2$, i.e. to the two integrals $$\int\frac{\sin x}{\cos^3x}\sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}\,dx\\ \int\frac{1}{\cos^2x}\sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}\,dx$$ – user228113 Aug 3 '16 at 16:01
• @Dr.SonnhardGraubner See it carefully . – Aakash Kumar Aug 3 '16 at 16:08
• @Dr.SonnhardGraubner I'd say that the poster tried some manipulations in order to get rid of cosecants, secants, a good part of the big radical and the dependence on $\alpha$, but he ultimately got discouraged when he could not recognise a sensible integrand despite the simplifications. – user228113 Aug 3 '16 at 16:15

$$I=\int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx,$$ $$I=\cos{\alpha}\int \frac{\sin x}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx+\sin{\alpha}\int \frac{1}{\cos^2 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx,$$ $$I=\cos{\alpha} I_1+\sin{\alpha}I_2,$$ where $$I_1=\int \tan{x}\sqrt{\frac{1+\tan{x}}{1-\tan{x}}} d(\tan{x}),$$ $$I_2=\int \sqrt{\frac{1+\tan{x}}{1-\tan{x}}}d(\tan{x}).$$ Substitute $t=\tan{x}$ in both integrals: $$I_1=\int t\sqrt{\frac{1+t}{1-t}}dt,$$ $$I_2=\int \sqrt{\frac{1+t}{1-t}}.$$

Let $$I = \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}dx$$

$$I = \cos \alpha\int \frac{\sin x+\cos x\cdot \tan \alpha}{\cos x}\cdot \sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$$

So $$I = \cos \alpha \int (\tan \alpha+\tan x)\sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$$

Now Put $\tan x= t\;,$ Then $\sec^2 xdx = dt$

So $$I = \cos \alpha \int (t+\tan \alpha)\sqrt{\frac{1+t}{1-t}}dt = \cos \alpha \int \frac{(t+\tan \alpha)(t+1)}{\sqrt{1-t^2}}dt$$

So $$I = \cos \alpha \int\frac{t^2+(\tan \alpha +1)t+\tan \alpha}{\sqrt{1-t^2}}dt$$

So $$I = \cos \alpha \int\frac{t^2}{\sqrt{1-t^2}}dt+\cos \alpha \int\frac{(\tan \alpha +1)}{\sqrt{1-t^2}}dt+\cos \alpha\int\frac{\tan \alpha}{\sqrt{1-t^2}}dt$$

In first put $t=\sin \phi\;,$ Then $dt = \cos \phi d\phi$ and in second $(1-t^2)=u^2\;,$ Then $tdt = -udu$

So $$I = \frac{\cos \alpha}{2} \int (1-\cos 2 \phi)d\phi-\cos \alpha (1+\tan \alpha)\int du+\sin \alpha \cdot \sin^{-1}(t)$$

So $$I = \frac{\cos \alpha}{2}\left[\sin^{-1}(t)-t\sqrt{1-t^2}\right]-\cos \alpha(1+\tan \alpha)\sqrt{1-t^2}+\sin\alpha\cdot \sin^{-1}(t)+\mathcal{C}$$

So $$I = \frac{\cos \alpha}{2}\left[\sin^{-1}(\tan x)-t\sqrt{1-\tan^2x}\right]-\cos \alpha(1+\tan \alpha)\sqrt{1-\tan^2x}+\sin \alpha \cdot \sin^{-1}(\tan x)+\mathcal{C}$$