#### Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 18 textbook solution.

Answer:-  $\frac{\pi}{12}$

Hints:-  You must know the integration rules of trigonometric functions and its limits.

Given:-  $\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\cot ^{3 / 2} x} d x$

Solution : $\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\cot ^{3 / 2} x} d x=I$                                             ...(1)

$I=\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\cot \left(\frac{\pi}{2}-x\right)^{3 / 2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(1+b-x) d x\right]$

\begin{aligned} &I=\int_{\pi / 6}^{\pi / 3} \frac{1}{\tan ^{3 / 2}+1} d x \\ &I=\int_{\pi / 6}^{\pi / 3} \frac{\cot ^{1 / 2} x}{1+\cot ^{3 / 2} x} d x \end{aligned}                                                               ...(2)

\begin{aligned} &2 I=\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\cot ^{3 / 2} x} d x+\int_{\pi / 6}^{\pi / 3} \frac{\cot ^{1 / 2} x}{1+\cot ^{3 / 2} x} d x \\ &2 I=\int_{\pi / 6}^{\pi / 3} \frac{1+\cot ^{1 / 2} x}{1+\cot ^{3 / 2} x} d x \\ &2 I=\int_{\pi / 6}^{\pi / 3} 1 \cdot d x \end{aligned}

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

$I=\frac{\pi}{12}$