Question

The integral $\int \sqrt{ 1 + 2 \cot \, x (cosec \, x + \cot \, x) } dx \, \left( 0 < x < \frac{\pi}{2} \right)$ is equal to : (where $C$ is a constant of integration)

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

Answer 1

The Correct Option is B

To solve the integral $\int \sqrt{1 + 2 \cot x (\csc x + \cot x)} \, dx$, we need to simplify the expression inside the square root.

Given expression inside the square root:

$1 + 2 \cot x (\csc x + \cot x)$

Simplifying, we have:

• $= 1 + 2 \cot x (\csc x + \cot x)$
• $= 1 + 2 \cot x (\frac{1}{\sin x} + \frac{\cos x}{\sin x})$
• $= 1 + 2 \frac{\cos x}{\sin x} \cdot \frac{1+\cos x}{\sin x}$
• $= 1 + 2 \frac{\cos x (1+\cos x)}{\sin^2 x}$
• $= 1 + 2 \frac{\cos x + \cos^2 x}{1 - \cos^2 x}$

This can be simplified further knowing that $1 - \cos^2 x = \sin^2 x$:

• $= \frac{\sin^2 x + 2\cos x(1 + \cos x)}{\sin^2 x}$
• $= \frac{\sin^2 x + 2\cos x + 2\cos^2 x}{\sin^2 x}$
• $= \frac{(1 + \cos x)^2}{\sin^2 x}$

This means the integrand becomes $\sqrt{\frac{(1 + \cos x)^2}{\sin^2 x}} = \frac{1 + \cos x}{\sin x}$.

This simplifies to:

$\frac{1 + \cos x}{\sin x} = \csc x + \cot x$

The integral thus becomes:

• $\int (\csc x + \cot x) \, dx$

We can integrate directly:

• $= \int \csc x \, dx + \int \cot x \, dx$
• The integral of $\csc x$ is $-\log\left| \csc x + \cot x \right|$.
• The integral of $\cot x$ is $\log | \sin x |$.

Thus, the integral becomes:

• $- \log\left| \csc x + \cot x \right| + \log | \sin x | + C$

Since $\csc x + \cot x = \frac{1 + \cos x}{\sin x}$, we can further simplify:

• $= \log | \sin x | - \log \left| \frac{1 + \cos x}{\sin x} \right| + C$
• $= \log | \sin^2 x | - \log | 1 + \cos x | + C$
• Using the identity $1 + \cos x = 2\cos^2\frac{x}{2}$, we have:
• This simplifies to:
• $= 2 \log | \sin x | - \log | 1 + \cos x | + C$
• $= 2 \log \left|\sin x\right| - \log | 2 \cos^2 \frac{x}{2} | + C$
• $= 2 \log \left|\sin x\right| - 2 \log \left|\cos\frac{x}{2}\right| + C$
• $= 2 \log \left( \frac{\sin \frac{x}{2}}{ \cos\frac{x}{2}} \right) + C$
• $= 2 \log \left(\sin \frac{x}{2}\right) + C$

Therefore, the integral is $2 \log\left(\sin \frac{x}{2}\right) + C$.