#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 66 Maths Textbook Solution.

Answer: $\frac{4}{\sqrt{3}}$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(\tan x+\cot x)^{2} d x$

Solution:$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(\tan x+\cot x)^{2} d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\tan ^{2} x+\cot ^{2} x+2 \tan x \cot x\right) d x\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

\begin{aligned} &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\left(\sec ^{2} x-1\right)+\left(\cos e c^{2} x-1\right)+2 \tan x \frac{1}{\tan x}\right) d x \\ &\left.=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\sec ^{2} x-1\right)+\left(\cos e c^{2} x-1\right)+2 \tan x \frac{1}{\tan x}\right) d x \\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{2} x d x-1 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x+\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos e c^{2} x d x-1 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x+2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x \\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\pi}{3} \sec ^{2} x d x+\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos e c^{2} x d x-2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x+2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x \end{aligned}

$\left[\begin{array}{l} \int \sec ^{2} x d x=\tan x \\ \int \cos e c^{2} x d x=-\cot x \end{array}\right]$

\begin{aligned} &=\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)-\left(\frac{1}{\sqrt{3}}-\sqrt{3}\right) \\ &=\left(\sqrt{3}-\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}+\sqrt{3}\right) \\ &=2 \sqrt{3}-\frac{2}{\sqrt{3}} \\ &=\frac{2 \times 3-2}{\sqrt{3}} \\ &=\frac{6-2}{\sqrt{3}} \\ &=\frac{4}{\sqrt{3}} \end{aligned}