Evaluate the following Integral:\(\int\limits_{\pi/6}^{\pi/3}\)(tan x

1.	Evaluate the following Integral:\(\int\limits_{\pi/6}^{\pi/3}\)(tan x + cot x)2 dx
Answer» =\(\int\limits_{\pi/6}^{\pi/3}\)tan²x + 2tanx cot x + cot²x dx recall: sec²x - tan²x = 1, cosec²x - cot²x = 1 = \(\int\limits_{\pi/6}^{\pi/3}\)sec²x -1 + 2 + cosec²x - 1 dx = \(\int\limits_{\pi/6}^{\pi/3}\) sec²x + cosec²x dx Integral sec²x is tan x and integral of cosec²x = - cot x = [tan x]\(\cfrac{\frac{\pi}3}{\frac{\pi}6}\) - [cot x]\(\cfrac{\frac{\pi}3}{\frac{\pi}6}\) Tan 30 = cot 60 = \(\cfrac{1}{\sqrt3}\) Tan 60 = cot 30 = √3 = Tan 60 - cot 60 - {Tan 30 - cot 30} = \(\sqrt3-\cfrac{1}{\sqrt3}-(\cfrac{1}{\sqrt3}-\sqrt3)\) = \(\cfrac{4}{\sqrt3}\)

Evaluate the following Integral:\(\int\limits_{\pi/6}^{\pi/3}\)(tan x + cot x)2 dx

Answer»

=\(\int\limits_{\pi/6}^{\pi/3}\)tan²x + 2tanx cot x + cot²x dx

recall: sec²x - tan²x = 1, cosec²x - cot²x = 1

= \(\int\limits_{\pi/6}^{\pi/3}\)sec²x -1 + 2 + cosec²x - 1 dx

= \(\int\limits_{\pi/6}^{\pi/3}\) sec²x + cosec²x dx

Integral sec²x is tan x and integral of cosec²x = - cot x

= [tan x]\(\cfrac{\frac{\pi}3}{\frac{\pi}6}\) - [cot x]\(\cfrac{\frac{\pi}3}{\frac{\pi}6}\)

Tan 30 = cot 60 = \(\cfrac{1}{\sqrt3}\)

Tan 60 = cot 30 = √3

= Tan 60 - cot 60 - {Tan 30 - cot 30}

= \(\sqrt3-\cfrac{1}{\sqrt3}-(\cfrac{1}{\sqrt3}-\sqrt3)\)

= \(\cfrac{4}{\sqrt3}\)