Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 12 maths

$\frac{\pi }{12}$

Given:

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} d x$

Hint:

Using $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

Explanation:

Let

\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} d x \\\\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\frac{\cos x}{\sin x}}} d x, \quad\left[\therefore \cot x=\frac{\cos x}{\sin x}\right] \end{aligned}

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \ldots(i)$

We have a property that

$\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

Using this, we get

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}+\sqrt{\sin \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}} d x$

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x,\left[\therefore \sin \left(\frac{\pi}{2}-x\right)=\cos x, \quad \cos \left(\frac{\pi}{2}-x\right)=\sin x\right]$

$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \ldots(i i)$

\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\\\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x \end{aligned}

\begin{aligned} &I=\frac{1}{2}[x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\\\ &I=\frac{1}{2}\left[\frac{\pi}{3}-\frac{\pi}{6}\right] \end{aligned}

\begin{aligned} &I=\frac{1}{2}\left[\frac{2 \pi-2 \pi}{6}\right] \\\\ &I=\frac{1}{2} \times \frac{\pi}{6} \\\\ &=\frac{\pi}{12} \end{aligned}