# Evaluation of $\int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$

Evaluation of $\displaystyle \int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$

$\bf{My \; Try::}$ Let $\displaystyle I = \int \frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx$

So $\displaystyle I = \int\frac{\sin (x+\alpha)}{\cos x}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx\cdot \sec^2 x$

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

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

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

$\displaystyle I = \int (\cos \alpha\cdot t + \sin \alpha)\sqrt{\frac{1+t}{1-t}}dt$

Now How can I solve after that

Help me

Thanks

## 2 Answers

Assuming what you have done up till the last part is correct, these hints I think will help you go further:

(1) Multiplying and dividing by $\sqrt{1+t}$

(2) Knowing the derivative of $\sinh^{-1} (\sqrt{t-1})$ and $\sqrt{t+1}\sqrt{t-1}$

(3) Using partial integration

Let \begin{align} I = \int \frac{\sin(x+a)}{\cos^3(x)}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx \end{align} then \begin{align} I &= \int (\sin(x) \cos(a) + \cos(x) \sin(a))(\cos(x) + \sin(x)) \ \frac{dx}{\cos^{3}(x) \sqrt{\cos(2x)}} \\ &= \cos(a) \ \int \frac{\sin^{2}(x) \ dx}{\cos^{3}(x) \sqrt{\cos(2x)}} + \sin(a) \ \int \frac{ dx}{\cos(x) \sqrt{\cos(2x)}} + (\cos(a) + \sin(a)) \ \int \frac{\sin(x) \ dx}{\cos^{2}(x) \sqrt{\cos(2x)}} \\ &= \frac{1}{2} (\cos(a) + 2 \sin(a)) \ \tan^{-1}\left( \frac{\sin(x)}{\sqrt{\cos(2x)}}\right) - \frac{1}{2} \cos(a) \sqrt{\cos(2x)} \ \frac{\sin(x)}{\cos^{2}(x)} - (\cos(a) + \sin(a)) \ \frac{\sqrt{\cos(2x)}}{\cos(x)} \end{align} where there integrals \begin{align} \int \frac{\sin^{2}(x) \ dx}{\cos^{3}(x) \sqrt{\cos(2x)}} &= \frac{1}{2} \left[ \tan^{-1}\left( \frac{\sin(x)}{\sqrt{\cos(2x)}}\right) - \frac{\sin(x) \ \sqrt{\cos(2x)}}{\cos^{2}(x)} \right] \\ \int \frac{\sin(x) \ dx}{\cos^{2}(x) \sqrt{\cos(2x)}} &= - \frac{ \sqrt{\cos(2x)}}{\cos^{2}(x)} \\ \int \frac{ dx}{\cos(x) \sqrt{\cos(2x)}} &= \tan^{-1}\left( \frac{\sin(x)}{\sqrt{\cos(2x)}}\right) \end{align} were used.