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: Understanding the Question:
We are asked to evaluate an indefinite integral containing an algebraic combination of trigonometric functions under a square root.

Step 2: Key Formula or Approach:
The key is to manipulate the expression inside the square root to form a perfect square.
Use the Pythagorean identity: $1 + \tan^2 x = \sec^2 x$.
Also, recall the standard integrals: $\int \sec x \, dx = \log|\sec x + \tan x| + c$ and $\int \tan x \, dx = \log|\sec x| + c$.

Step 3: Detailed Explanation:
Let $I = \int[1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} d x$.
First, expand the expression inside the bracket:
$1 + 2\tan^2 x + 2\tan x \sec x$
Rewrite $2\tan^2 x$ as $\tan^2 x + \tan^2 x$:
$1 + \tan^2 x + \tan^2 x + 2\tan x \sec x$
Substitute $\sec^2 x$ for $(1 + \tan^2 x)$:
$\sec^2 x + \tan^2 x + 2\tan x \sec x$
Notice that this is a perfect square expansion of the form $a^2 + b^2 + 2ab$:
$(\sec x + \tan x)^2$
Now substitute this back into the integral:
$$I = \int \left[ (\sec x + \tan x)^2 \right]^{\frac{1}{2}} dx$$ $$I = \int (\sec x + \tan x) dx$$ Break the integral into two parts:
$$I = \int \sec x \, dx + \int \tan x \, dx$$ Evaluate the standard integrals:
$$I = \log|\sec x + \tan x| + \log|\sec x| + c$$ Using the logarithmic property $\log m + \log n = \log(mn)$, we can combine the terms:
$$I = \log|\sec x (\sec x + \tan x)| + c$$

Step 4: Final Answer:
The integrated expression is $\log[\sec x(\sec x + \tan x)] + c$, corresponding to option (C).