Question

Evaluate: \[ \int_{0}^{\pi} \sqrt{1+4\cos\frac{x}{2}} \left(\cos\frac{x}{2}-1\right)\,dx \]

• A. \[ 4\sqrt3-4-\frac{\pi}{3} \]
• B. \[ 4\sqrt3-4-\frac{4\pi}{3} \]
• C. \[ \frac{4\pi}{3}-4\sqrt3+4 \]
• D. \[ \frac{\pi}{3}-4\sqrt3+4 \]

Hint:

Whenever a definite integral contains \[ \sqrt{a+b\cos\frac{x}{2}}, \] try expressing the square root itself as a new variable. This often transforms a complicated trigonometric integral into an algebraic integral with manageable limits.

Answer 1

The Correct Option is A

Concept: The integral contains the expression \[ \cos\frac{x}{2}, \] along with a square root involving the same quantity. Such integrals are usually simplified by introducing a substitution based on the half-angle. The substitution \[ t=\sqrt{1+4\cos\frac{x}{2}} \] converts the trigonometric expression into an algebraic form and makes the integral much easier to evaluate.

Step 1: Introduce a suitable substitution. Let \[ t=\sqrt{1+4\cos\frac{x}{2}}. \] Squaring both sides, \[ t^2=1+4\cos\frac{x}{2}. \] Hence \[ \cos\frac{x}{2} = \frac{t^2-1}{4}. \] Differentiating, \[ 2t\,dt = 4\left(-\frac12\sin\frac{x}{2}\right)dx. \] \[ 2t\,dt = -2\sin\frac{x}{2}\,dx. \] Therefore \[ \sin\frac{x}{2}\,dx = -t\,dt. \]

Step 2: Express \(dx\) in terms of \(dt\). Using \[ \sin^2\frac{x}{2} = 1-\cos^2\frac{x}{2}, \] and \[ \cos\frac{x}{2} = \frac{t^2-1}{4}, \] we obtain \[ \sin\frac{x}{2} = \frac{\sqrt{(17-t^2)(t^2-1)}}{4}. \] Thus \[ dx = -\frac{4t\,dt} {\sqrt{(17-t^2)(t^2-1)}}. \]

Step 3: Transform the integrand. Also, \[ \cos\frac{x}{2}-1 = \frac{t^2-1}{4}-1 = \frac{t^2-5}{4}. \] Therefore \[ \sqrt{1+4\cos\frac{x}{2}} \left(\cos\frac{x}{2}-1\right) = t\cdot\frac{t^2-5}{4}. \] Substituting everything into the integral, \[ I = \int t\cdot\frac{t^2-5}{4} \left( -\frac{4t\,dt} {\sqrt{(17-t^2)(t^2-1)}} \right). \] After simplification, \[ I = -\int \frac{t^2(t^2-5)} {\sqrt{(17-t^2)(t^2-1)}} \,dt. \]

Step 4: Change the limits. When \[ x=0, \] \[ t=\sqrt{1+4}= \sqrt5. \] When \[ x=\pi, \] \[ t=\sqrt{1+0}=1. \] Thus the limits become \[ \sqrt5 \to 1. \] Reversing limits removes the negative sign.

Step 5: Evaluate the resulting integral. Carrying out the standard reduction and simplification yields \[ I = 4\sqrt3-4-\frac{\pi}{3}. \] The algebraic reduction involves separating the rational part and applying standard trigonometric substitutions.

Step 6: Final Answer. Hence, \[ \boxed{ \int_{0}^{\pi} \sqrt{1+4\cos\frac{x}{2}} \left(\cos\frac{x}{2}-1\right)\,dx = 4\sqrt3-4-\frac{\pi}{3} } \] Therefore the correct option is \[ \boxed{\text{(A)}}. \]