Question

Evaluate: \[ \int \frac{\cos 2x+\sin 4x} {\sqrt{3\sin 2x-2}} \,dx \]

• A. \[ \frac1{27} \sqrt{3\sin2x-2} \,(6\sin2x+17)+C \]
• B. \[ \frac{\sqrt{3\sin2x-2}} {27(6\sin2x+17)} +C \]
• C. \[ \frac{27(6\sin2x+17)} {\sqrt{3\sin2x-2}} +C \]
• D. \[ 27\sqrt{3\sin2x-2} (6\sin2x+17)+C \]

Hint:

Whenever an expression of the form \[ \sqrt{a+b\sin 2x} \] appears, try substituting the quantity inside the square root. The resulting integral usually reduces to a simple power integral.

Answer 1

The Correct Option is A

Concept: The expression under the square root, \[ 3\sin2x-2, \] suggests the substitution \[ t=3\sin2x-2. \] The numerator can then be manipulated so that it contains \(dt\).

Step 1: Simplify the numerator. Using \[ \sin4x=2\sin2x\cos2x, \] we get \[ \cos2x+\sin4x = \cos2x(1+2\sin2x). \] Therefore \[ I = \int \frac{\cos2x(1+2\sin2x)} {\sqrt{3\sin2x-2}} dx. \]

Step 2: Use substitution. Let \[ t=3\sin2x-2. \] Differentiating, \[ dt = 6\cos2x\,dx. \] Hence \[ \cos2x\,dx = \frac{dt}{6}. \] Also, \[ \sin2x=\frac{t+2}{3}. \] Substituting, \[ I = \frac16 \int \frac{1+\frac{2(t+2)}3} {\sqrt t} \,dt. \] \[ = \frac1{18} \int \frac{2t+7} {\sqrt t} \,dt. \]

Step 3: Rewrite the integrand. \[ I = \frac1{18} \int \left( 2t^{1/2} + 7t^{-1/2} \right) dt. \] Integrating term-by-term, \[ I = \frac1{18} \left[ \frac43t^{3/2} + 14t^{1/2} \right] +C. \] \[ = \frac1{27} t^{1/2}(2t+21)+C. \]

Step 4: Substitute back. Since \[ t=3\sin2x-2, \] \[ 2t+21 = 6\sin2x+17. \] Therefore \[ I = \frac1{27} \sqrt{3\sin2x-2} \, (6\sin2x+17) +C. \]

Step 5: Final Answer. \[ \boxed{ \frac1{27} \sqrt{3\sin2x-2} (6\sin2x+17) +C } \] Hence option (A) is correct.