Question

$\int \sqrt{\frac{1+2\tan x}{(\sec x + \tan x)}} dx$

• A. $\int \sqrt{\sec x + \tan x} dx$
• B. $\int \sqrt{\sec x - \tan x} dx$
• C. $\int \sqrt{\frac{\cos x + 2\sin x}{1+\sin x}} dx$
• D. $\int \sqrt{\frac{1-2\tan x}{\sec x - \tan x}} dx$

Answer 1

Answer: (C)

$\int \sqrt{\frac{\cos x + 2\sin x}{1+\sin x}} dx$

Answer 2

To simplify the integral $\int \sqrt{\frac{1+2\tan x}{(\sec x + \tan x)}} dx$, we first rewrite the trigonometric functions in terms of $\sin x$ and $\cos x$: $\tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x}$ Substitute these into the expression inside the square root: $\frac{1+2\tan x}{\sec x + \tan x} = \frac{1+2\frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}$ Multiply the numerator and the denominator by $\cos x$: $= \frac{\cos x (1+2\frac{\sin x}{\cos x})}{\cos x (\frac{1}{\cos x} + \frac{\sin x}{\cos x})} = \frac{\cos x + 2\sin x}{1+\sin x}$ Therefore, the integral can be rewritten as: $\int \sqrt{\frac{\cos x + 2\sin x}{1+\sin x}} dx$