Question

Evaluate \[ \int \sqrt{1+2\cot x(\cot x+\cosec x)}\,dx \]

• A. \(2\log\left(\sin\frac{x}{2}\right)+C\)
• B. \(2\log(\sin x)-\log(\cosec x+\cot x)+C\)
• C. \(\frac{1}{2}\log\left(\cosec\frac{x}{2}+\cot\frac{x}{2}\right)+C\)
• D. \(4\log\cos\frac{x}{2}+C\)

Hint:

Use the identity \(\cosec x+\cot x=\cot\frac{x}{2}\) whenever an integral contains the expression \(\cosec x+\cot x\).

Answer 1

The Correct Option is A

Step 1: Simplify the expression inside the square root.
We have \[ 1+2\cot x(\cot x+\cosec x) \] \[ =1+2\cot^2x+2\cot x\cosec x \] Using \[ \cosec^2x=1+\cot^2x, \] we get \[ 1+2\cot^2x+2\cot x\cosec x = \cosec^2x+\cot^2x+2\cot x\cosec x \] \[ =(\cosec x+\cot x)^2 \] Therefore, \[ \sqrt{1+2\cot x(\cot x+\cosec x)} = \cosec x+\cot x \]

Step 2: Convert into half-angle form.
We know that \[ \cosec x+\cot x=\cot\frac{x}{2} \] So, \[ \int \sqrt{1+2\cot x(\cot x+\cosec x)}\,dx = \int \cot\frac{x}{2}\,dx \]

Step 3: Integrate.
Let \[ u=\frac{x}{2} \] Then, \[ dx=2du \] Hence, \[ \int \cot\frac{x}{2}\,dx = 2\int \cot u\,du \] \[ =2\log(\sin u)+C \] Substituting back, \[ =2\log\left(\sin\frac{x}{2}\right)+C \]

Step 4: Final conclusion.
Therefore, \[ \boxed{2\log\left(\sin\frac{x}{2}\right)+C} \]