Question

How do you find the integral of $\int {{{\tan }^8}} x.{\sec ^2}xdx$?

Answer 1

Hint: According to the given question, we have to find the integral of $ \int {{{\tan }^8}} x.{\sec ^2}xdx $ . So, first of all we have to use the substitution technique in which we have to let the $ \tan x $ equals to any variable like $ a $ or $ m $ or $ z $ etc.
Now, we have to differentiate $ \tan x $ with respect to $ x $ from both sides and obtain the value of $ dx $ in terms of that variable.
Now, we have to put all the values of $ \tan x $ and $ dx $ in the given expression $ \int {{{\tan }^8}} x.{\sec ^2}xdx $ .
Now, we have to integrate that expression obtained after putting the values of $ \tan x $ and $ dx $ with the help of the formula as mentioned below.
 $ \Rightarrow \int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C...................................(A) $

Complete step-by-step solution:
Step 1: First of all we have to let the $ \tan x $ equals to any variable like $ t $ as mentioned below,
 $ \Rightarrow \tan x = t........................(1) $
Step 2: Now, we have to differentiate the expression (1) as obtained in the solution step 1 with respect to $ x $ from both sides.
 $ \Rightarrow \dfrac{d}{{dx}}\left( {\tan x} \right) = \dfrac{d}{{dx}}\left( t \right) $
Now, as we know that the differentiation of $ \tan x $ is equal to $ {\sec ^2}x $ .
 $
   \Rightarrow {\sec ^2}x = \dfrac{{dt}}{{dx}} \\
   \Rightarrow {\sec ^2}xdx = dt...........................(2)
 $
Step 3: Now, we have to put the values of the expression (1) and (2) as obtain in the solution step 1 and solution step 2 respectively in the given expression in the question as $ \int {{{\tan }^8}} x.{\sec ^2}xdx $ .
 $ \Rightarrow \int {{t^8}} dt $
Now, we have to differentiate the expression obtained just above by using the formula (A) as mentioned in the solution hint.
 $
   \Rightarrow \dfrac{{{t^{8 + 1}}}}{{8 + 1}} + C \\
   \Rightarrow \dfrac{{{t^9}}}{9} + C
 $
Step 4: Now, we have to put the value of $ t $ as $ \tan x $ in the expression obtained in the solution step 3.
 $ \Rightarrow \dfrac{{{{\tan }^9}x}}{9} + C $

Hence, the integral of the given expression as $ \int {{{\tan }^8}} x.{\sec ^2}xdx $ is $ \dfrac{{{{\tan }^9}x}}{9} + C $

Note: It is necessary to let $ \tan x $ equals to any variable like $ a $ or $ m $ or $ z $ etc.
It is necessary to differentiate $ \tan x $ with respect to $ x $ from both sides and obtain the value of $ dx $ in terms of that variable.