Question

$\int\frac{\sin 2x}{\sin x}dx=$ ________.

• A. $\sin x+C$
• B. $2 \cos x+C$
• C. $-\cos x+C$
• D. $-\sin x+C$
• E. $2 \sin x+C$

Hint:

Always simplify trigonometric fractions before integrating.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
We need to find the indefinite integral of the given trigonometric function. The best approach is to first simplify the integrand using trigonometric identities before performing the integration.
Step 2: Key Formula or Approach
We will use the double-angle identity for sine:
\[ \sin(2x) = 2\sin x \cos x \] After simplifying the integrand, we will use the basic integration formula \(\int \cos x \, dx = \sin x + C\).
Step 3: Detailed Explanation
1. Simplify the integrand.
The integrand is \(\frac{\sin 2x}{\sin x}\).
Substitute the double-angle identity into the numerator:
\[ \frac{2\sin x \cos x}{\sin x} \] Assuming \(\sin x \neq 0\), we can cancel the \(\sin x\) term.
\[ 2\cos x \] 2. Integrate the simplified expression.
The integral becomes:
\[ \int 2\cos x \, dx \] We can pull the constant 2 out of the integral:
\[ 2 \int \cos x \, dx \] Now, we use the standard integral of cosine:
\[ 2 (\sin x) + C \] \[ 2\sin x + C \] Step 4: Final Answer
The integral is \(2\sin x + C\).