Question

The integral \(\rm \int \sqrt {1 + 2 \cot x (cosec x + \cot x)} dx,\; \left( 0 < x < \frac \pi 2 \right)\) (where C is a constant of integration) is equal to
1. \(2 \log \left(\sin \frac x 2\right) + C\)
2. \(2 \log \left(\cos \frac x 2\right) + C\)
3. \(4 \log \left(\cos \frac x 2\right) + C\)
4. \(4 \log \left(\sin \frac x 2\right) + C\)

Answer 1

Correct Answer - Option 1 : \(2 \log \left(\sin \frac x 2\right) + C\)

Concept:

\(\rm \int \cot x dx = \log \sin x + c\)

cosec ² x - cot² x = 1

1 + cos x = 2cos² \(\rm \frac x 2\)

Calculation:

I = \(\rm \int \sqrt {1 + 2 \cot x (cosec x + \cot x)} dx,\; \left( 0 < x < \frac \pi 2 \right)\)

= \(\rm \int \sqrt {cosec^2\;x - \cot^2 x+ 2 \cot x cosec \;x + 2\cot^2 x} dx\)

= \(\rm \int \sqrt {cosec^2\;x + \cot^2 x+ 2 \cot x cosec \;x } dx\)

= \(\rm \int \sqrt{(cosec \;x + \cot x)^2} \; dx\)

= \(\rm \int cosec \;x + \cot x \; dx\)

= \(\rm \int \left(\frac{1}{\sin x} + \frac{\cos x}{\sin x} \right ) dx\)

= \(\rm \int \frac{1+\cos x}{\sin x} dx\)

= \(\rm \int \frac{2 \cos^2 \frac{x}{2}}{2\sin \frac{x}{2} \cos \frac{x}{2}} dx\)

= \(\rm \int \cot \frac{x}{2} dx\)

Let \(\rm \frac x 2 =t\)

⇒ x = 2t

Differentiating with respect to x, we get

⇒ dx = 2dt

Now, 

I = \(\rm 2\int \cot t dt\)

= \(\rm 2 \log \left(\sin t\right) + C\)

= \(2 \log \left(\sin \frac x 2\right) + C\)