Question

\(\int \left(\frac{\sin 3x}{\sin x}-\frac{\cos 3x}{\cos x}\right)\,dx\) is equal to

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

Hint:

Whenever \(\sin 3x\) or \(\cos 3x\) appears with division by \(\sin x\) or \(\cos x\), use triple-angle formulas first. The expression usually simplifies dramatically.

Answer 1

The Correct Option is D

Step 1: Write the integral clearly.
We need to evaluate:
\[ I=\int \left(\frac{\sin 3x}{\sin x}-\frac{\cos 3x}{\cos x}\right)\,dx \]

Step 2: Use the triple-angle formulas.
Recall: \[ \sin 3x=3\sin x-4\sin^3 x \] and \[ \cos 3x=4\cos^3 x-3\cos x \]

Step 3: Divide each expression by \(\sin x\) and \(\cos x\).
Then, \[ \frac{\sin 3x}{\sin x}=\frac{3\sin x-4\sin^3 x}{\sin x}=3-4\sin^2 x \] and \[ \frac{\cos 3x}{\cos x}=\frac{4\cos^3 x-3\cos x}{\cos x}=4\cos^2 x-3 \]

Step 4: Substitute back into the integrand.
So the integrand becomes: \[ (3-4\sin^2 x)-(4\cos^2 x-3) \] \[ =3-4\sin^2 x-4\cos^2 x+3 \] \[ =6-4(\sin^2 x+\cos^2 x) \]

Step 5: Use the identity \(\sin^2 x+\cos^2 x=1\).
Therefore, \[ 6-4(\sin^2 x+\cos^2 x)=6-4(1)=2 \] So the integral reduces to: \[ I=\int 2\,dx \]

Step 6: Integrate the constant.
\[ I=2x+C \]

Step 7: State the final answer.
Hence, \[ \boxed{\int \left(\frac{\sin 3x}{\sin x}-\frac{\cos 3x}{\cos x}\right)\,dx=2x+C} \] which matches option \((4)\).