2
$\begingroup$

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

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

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

$\endgroup$
  • $\begingroup$ 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 $$\int\frac{\sin x}{\cos^3x}\sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}\,dx\\ \int\frac{1}{\cos^2x}\sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}\,dx$$ $\endgroup$ – user228113 Aug 3 '16 at 16:01
  • $\begingroup$ @Dr.SonnhardGraubner See it carefully . $\endgroup$ – Aakash Kumar Aug 3 '16 at 16:08
  • $\begingroup$ @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. $\endgroup$ – user228113 Aug 3 '16 at 16:15
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.