Question

Evaluate: \[ \int \frac{\sin4x}{\sin x}\,dx \]

• A. \(4(3\sin x+3\sin3x)+C\)
• B. \(\frac43(2\sin^3x+3\sin x)+C\)
• C. \(4(3\sin x-3\sin x)+C\)
• D. \(\frac43(3\sin x-2\sin^3x)+C\)

Hint:

Whenever \(\sin nx\) appears divided by \(\sin x\), first use multiple-angle identities. This often converts the problem into a simple polynomial integral after substitution.

Answer 1

The Correct Option is D

Concept: To integrate expressions involving \(\sin4x\), it is often useful to express the multiple-angle function in terms of powers of \(\sin x\) and \(\cos x\). The identity \[ \sin4x = 4\sin x\cos x\cos2x \] helps simplify the integrand considerably.

Step 1: Simplify the integrand. Using \[ \sin4x = 4\sin x\cos x\cos2x, \] we obtain \[ \frac{\sin4x}{\sin x} = 4\cos x\cos2x. \] Hence \[ I = \int4\cos x\cos2x\,dx. \]

Step 2: Express \(\cos2x\) in terms of \(\sin x\). Using \[ \cos2x = 1-2\sin^2x, \] we get \[ I = 4\int\cos x(1-2\sin^2x)\,dx. \] \[ = 4\int(\cos x-2\sin^2x\cos x)\,dx. \]

Step 3: Use substitution. Let \[ t=\sin x. \] Then \[ dt=\cos x\,dx. \] Thus \[ I = 4\int(1-2t^2)\,dt. \] \[ = 4\left(t-\frac23t^3\right)+C. \]

Step 4: Substitute back. Replacing \(t\) by \(\sin x\), \[ I = 4\left(\sin x-\frac23\sin^3x\right)+C. \] \[ = \frac43(3\sin x-2\sin^3x)+C. \]

Step 5: Final Answer. \[ \boxed{ \frac43(3\sin x-2\sin^3x)+C } \] Hence option (D) is correct.