Integrate $\int{\sec x\cdot \log \left( \sec x+\tan x \right)dx}$
Answer
585.3k+ views
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

The chemical formula of tear gas is A CO Cl 2 B C 10 class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

