
Integrate $\int{\sec x\cdot \log \left( \sec x+\tan x \right)dx}$
Answer
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Hint: In this question we have been given with an expression and we have to find the integration of the term. We will use the substitution method to solve the integration. We will substitute the term $\log \left( \sec x+\tan x \right)$ as $t$ and then simplify the expression to get $dx$ in the form of $dt$. We will then integrate the expression and then substitute the value of $t$ in the expression to get the required solution.
Complete step by step solution:
We have the expression given to us as:
$\Rightarrow \int{\sec x\cdot \log \left( \sec x+\tan x \right)dx}\to \left( 1 \right)$
Now consider:
$\Rightarrow \log \left( \sec x+\tan x \right)=t\to \left( 2 \right)$
Now we will differentiate both the sides. We can see that the function in the left-hand side is a composite function therefore, we will use chain rule.
We know that $\dfrac{d}{dx}\log x=\dfrac{1}{x}$ and derivative of $t=dt$ therefore, we get:
$\Rightarrow \dfrac{1}{\sec x+\tan x}\cdot \dfrac{d}{dx}\left( \sec x+\tan x \right)dx=dt$
Now we know that $\dfrac{d}{dx}\sec x=\sec x\tan x$ and $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$therefore, on substituting, we get:
$\Rightarrow \dfrac{\sec x\tan x+{{\sec }^{2}}x}{\sec x+\tan x}dx=dt$
On taking $\sec x$ common from the numerator, we get:
$\Rightarrow \dfrac{\sec x\left( \tan x+\sec x \right)}{\sec x+\tan x}dx=dt$
On cancelling the similar terms, we get:
$\Rightarrow \sec xdx=dt\to \left( 3 \right)$
Now on substituting equation $\left( 2 \right)$ and $\left( 3 \right)$ in equation $\left( 1 \right)$, we get:
\[ \int{tdt}\]
Now we know that $\int{t}=\dfrac{{{t}^{2}}}{2}+c$ therefore, on substituting, we get:
$= \dfrac{{{t}^{2}}}{2}+c$
On substituting the value of $t$ from equation $\left( 2 \right)$, we get:
$= \dfrac{\log {{\left( \sec x+\tan x \right)}^{2}}}{2}+c$, which is the required solution.
Note: In this question we have used the substitution method to solve. Substitution method should be used when there are terms in multiplication which represent the derivative of another term. The complex term should be substituted as $t$ and its derivative should be taken and rearranged such that terms in the integration can be substituted.
Complete step by step solution:
We have the expression given to us as:
$\Rightarrow \int{\sec x\cdot \log \left( \sec x+\tan x \right)dx}\to \left( 1 \right)$
Now consider:
$\Rightarrow \log \left( \sec x+\tan x \right)=t\to \left( 2 \right)$
Now we will differentiate both the sides. We can see that the function in the left-hand side is a composite function therefore, we will use chain rule.
We know that $\dfrac{d}{dx}\log x=\dfrac{1}{x}$ and derivative of $t=dt$ therefore, we get:
$\Rightarrow \dfrac{1}{\sec x+\tan x}\cdot \dfrac{d}{dx}\left( \sec x+\tan x \right)dx=dt$
Now we know that $\dfrac{d}{dx}\sec x=\sec x\tan x$ and $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$therefore, on substituting, we get:
$\Rightarrow \dfrac{\sec x\tan x+{{\sec }^{2}}x}{\sec x+\tan x}dx=dt$
On taking $\sec x$ common from the numerator, we get:
$\Rightarrow \dfrac{\sec x\left( \tan x+\sec x \right)}{\sec x+\tan x}dx=dt$
On cancelling the similar terms, we get:
$\Rightarrow \sec xdx=dt\to \left( 3 \right)$
Now on substituting equation $\left( 2 \right)$ and $\left( 3 \right)$ in equation $\left( 1 \right)$, we get:
\[ \int{tdt}\]
Now we know that $\int{t}=\dfrac{{{t}^{2}}}{2}+c$ therefore, on substituting, we get:
$= \dfrac{{{t}^{2}}}{2}+c$
On substituting the value of $t$ from equation $\left( 2 \right)$, we get:
$= \dfrac{\log {{\left( \sec x+\tan x \right)}^{2}}}{2}+c$, which is the required solution.
Note: In this question we have used the substitution method to solve. Substitution method should be used when there are terms in multiplication which represent the derivative of another term. The complex term should be substituted as $t$ and its derivative should be taken and rearranged such that terms in the integration can be substituted.
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