Question

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

Answer 1

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.