Question

$\int[1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} d x=$

• A. $\log[\sec x(\sec x - \tan x)] + c$
• B. $\log[\csc x(\sec x + \tan x)] + c$
• C. $\log[\sec x(\sec x + \tan x)] + c$
• D. $\log[\sec x + \tan x] + c$

Hint:

Whenever you see $1$ and $\tan^2 x$ in an integrand, immediately try to convert them into $\sec^2 x$ to simplify the expression, especially if they are trapped under a radical!

Answer 1

The Correct Option is C

Step 1: Tidy the inside of the root.
Expand $1 + 2\tan x(\tan x + \sec x) = 1 + 2\tan^2 x + 2\tan x\sec x$.

Step 2: Make a perfect square.
Split $2\tan^2 x$ and use $1 + \tan^2 x = \sec^2 x$ to get $\sec^2 x + \tan^2 x + 2\tan x\sec x = (\sec x + \tan x)^2$.

Step 3: Integrate.
The square root removes the square, so we integrate $\sec x + \tan x$.
$$\int(\sec x + \tan x)\,dx = \log|\sec x + \tan x| + \log|\sec x| + c = \log[\sec x(\sec x + \tan x)] + c$$
\[ \boxed{\log[\sec x(\sec x + \tan x)] + c} \]