Question

Evaluate $\int \frac{\sec x}{\sqrt{\left\{sin\right(2x+\alpha )+\sin \alpha \}}}dx$

• A. $\sqrt{2}{\left[(\tan x+\tan \alpha )\sec \alpha \right]}^{1/2}+C$
• B. $\sqrt{2}{\left[(\tan x-\tan \alpha )cot\alpha \right]}^{1/2}$
• C. $\sqrt{2}{\left[(\tan x-\tan \alpha )\sec \alpha \right]}^{1/2}$
• D. $\sqrt{2}{\left[(\tan x+\tan \alpha )\sec \alpha \right]}^{1/2}$

Answer 1

Answer: (A)

$\sqrt{2}{\left[(\tan x+\tan \alpha )\sec \alpha \right]}^{1/2}+C$

$\sin (2x+\alpha )+\sin \alpha =2\sin (x+\alpha )\cos x=2\cos x[\sin x\cos \alpha +\cos x\sin \alpha ]$

$=2{\cos }^{2}x[\tan x\cos \alpha +\sin \alpha ]$

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

Put $\tan x\cos \alpha +\sin \alpha =t$

$\therefore {\sec }^{2}x\cos \alpha dx=dt$ $\therefore I=\frac{1}{\sqrt{2}\cos \alpha }\int \frac{dt}{\sqrt{t}}=\frac{1}{\sqrt{2}\cos \alpha }2\sqrt{t}=\sqrt{2}{\left[\frac{\tan x\cos \alpha +\sin \alpha }{{\cos }^2\alpha }\right]}^{1/2}$ $=\sqrt{2}{[\tan x\sec \alpha +\sec \alpha \tan \alpha ]}^{1/2}+c$ $=\sqrt{2}{\left[(\tan x+\tan \alpha )\sec \alpha \right]}^{1/2}+c$