Question

If \[ \int \frac{\sin x + \cos x}{\sqrt{1 + 2 \sin x \cos x}} \, dx = \varphi(x) + C, \text{ then } \varphi(x) = \]

• A. \( \log x \)
• B. \( x \)
• C. \( \log (\sin x + \cos x) \)
• D. \( \log \sin (\cos x) \)

Hint:

For integrals involving trigonometric functions, try using standard trigonometric identities and substitution to simplify the integral before solving.

Answer 1

The Correct Option is B

Step 1: Simplify the integrand.
We are given the integral:
\[ \int \frac{\sin x + \cos x}{\sqrt{1 + 2 \sin x \cos x}} \, dx \]
First, simplify the expression inside the square root. Recall the trigonometric identity:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, the expression inside the square root becomes:
\[ 1 + 2 \sin x \cos x = 1 + \sin 2x \]
So the integral becomes:
\[ \int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx \]

Step 2: Use substitution.
Let \( u = \sin x + \cos x \). Then, differentiate \( u \) with respect to \( x \):
\[ du = (\cos x - \sin x) \, dx \]
So the integral simplifies to:
\[ \int \frac{du}{\sqrt{1 + u^2}} = \log (u + \sqrt{1 + u^2}) + C \]

Step 3: Conclusion.
Since \( u = \sin x + \cos x \), we have:
\[ \varphi(x) = x \]
Therefore, the correct answer is option (B), \( \varphi(x) = x \).