What is the integral of $\sec \left( x \right)$ ?
Answer
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Hint: To find the integral of $\sec \left( x \right)$ , we have to multiply and divide $\sec \left( x \right)$ by $\sec x+\tan x$ . Then we have to simplify the expression. Then we have to consider the numerator as u and differentiate u. Then we have to substitute these values in the expression to be integrated. Then, we have to integrate this function and substitute for u.
Complete step by step solution:
We have to find the integral of $\sec \left( x \right)$ . For this, let us multiply and divide $\sec \left( x \right)$ by $\sec x+\tan x$ .
$\Rightarrow \int{\sec \left( x \right)dx}=\int{\dfrac{\sec \left( x \right)\times \left( \sec x+\tan x \right)}{\sec x+\tan x}dx}$
Let us simplify the numerator by multiplying $\sec \left( x \right)$ with $\sec x+\tan x$ .
$\Rightarrow \int{\dfrac{{{\sec }^{2}}x+\sec x\tan x}{\sec x+\tan x}dx}...\left( i \right)$
Now, let us consider $u=\sec x+\tan x...\left( ii \right)$ . We have to differentiate equation (i). We know that $\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x$ and $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$ . Therefore, we can write the differentiation of u as
$du=\left( \sec x\tan x+{{\sec }^{2}}x \right)dt...\left( iii \right)$
Let us substitute (ii) and (iii) in (i).
$\Rightarrow \int{\dfrac{du}{u}}$
We know that $\int{\dfrac{dx}{x}}=\log \left| x \right|+C$ . Therefore, we can write the above integration as
$\Rightarrow \log \left| u \right|+C$
Let us substitute the value of u from equation (ii) in the above value.
$\Rightarrow \log \left| \sec x+\tan x \right|+C$
Hence, the integral of $\sec \left( x \right)$ is $\log \left| \sec x+\tan x \right|+C$ .
Note: Students must never forget to substitute back the value of u. They must not forget to add the constant C to the value after integration. They must know the integration of basic functions. Students must also the how to differentiate and the differentiation of basic functions. They may make mistakes by confusing the integration and differentiation of $\sec \left( x \right)$ . They may write $\int{\sec x}=\sec x\tan x$ . Students have a chance of making mistake when writing the differentiation of tan x as $\sec x\tan x$ .
Complete step by step solution:
We have to find the integral of $\sec \left( x \right)$ . For this, let us multiply and divide $\sec \left( x \right)$ by $\sec x+\tan x$ .
$\Rightarrow \int{\sec \left( x \right)dx}=\int{\dfrac{\sec \left( x \right)\times \left( \sec x+\tan x \right)}{\sec x+\tan x}dx}$
Let us simplify the numerator by multiplying $\sec \left( x \right)$ with $\sec x+\tan x$ .
$\Rightarrow \int{\dfrac{{{\sec }^{2}}x+\sec x\tan x}{\sec x+\tan x}dx}...\left( i \right)$
Now, let us consider $u=\sec x+\tan x...\left( ii \right)$ . We have to differentiate equation (i). We know that $\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x$ and $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$ . Therefore, we can write the differentiation of u as
$du=\left( \sec x\tan x+{{\sec }^{2}}x \right)dt...\left( iii \right)$
Let us substitute (ii) and (iii) in (i).
$\Rightarrow \int{\dfrac{du}{u}}$
We know that $\int{\dfrac{dx}{x}}=\log \left| x \right|+C$ . Therefore, we can write the above integration as
$\Rightarrow \log \left| u \right|+C$
Let us substitute the value of u from equation (ii) in the above value.
$\Rightarrow \log \left| \sec x+\tan x \right|+C$
Hence, the integral of $\sec \left( x \right)$ is $\log \left| \sec x+\tan x \right|+C$ .
Note: Students must never forget to substitute back the value of u. They must not forget to add the constant C to the value after integration. They must know the integration of basic functions. Students must also the how to differentiate and the differentiation of basic functions. They may make mistakes by confusing the integration and differentiation of $\sec \left( x \right)$ . They may write $\int{\sec x}=\sec x\tan x$ . Students have a chance of making mistake when writing the differentiation of tan x as $\sec x\tan x$ .
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