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Question: Evaluate the value of \(\int {x{{\sec }^2}xdx} \) ....

Evaluate the value of ∫xsec⁡2xdx\int {x{{\sec }^2}xdx} .

Explanation

Solution

Firstly, let ∫xsec⁡2xdx=I\int {x{{\sec }^2}xdx} = I .
Then, find the value of integral using the formula ∫f(x)g(x)dx=f(x)∫g(x)dx−∫[ddxf(x)⋅∫g(x)dx]dx\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right) \cdot \int {g\left( x \right)dx} } \right]dx} , here f(x)=xf\left( x \right) = x and g(x)=sec⁡2xg\left( x \right) = {\sec ^2}x .
Thus, find the value of I and hence we found the required answer.

Complete step-by-step answer:
Here, we are asked to find the value of ∫xsec⁡2xdx\int {x{{\sec }^2}xdx} .
Let ∫xsec⁡2xdx=I\int {x{{\sec }^2}xdx} = I .
⇒I=∫xsec⁡2xdx\Rightarrow I = \int {x{{\sec }^2}xdx}
Since, we know that the integral of the product of two functions f(x)f\left( x \right) and g(x)g\left( x \right) is ∫f(x)g(x)dx=f(x)∫g(x)dx−∫[ddxf(x)⋅∫g(x)dx]dx\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right) \cdot \int {g\left( x \right)dx} } \right]dx} .
So, we will find the integral of I, by using the above formula, where f(x)=xf\left( x \right) = x and g(x)=sec⁡2xg\left( x \right) = {\sec ^2}x .
⇒I=x⋅∫sec⁡2xdx−∫[dxdx⋅∫sec⁡2xdx]dx\Rightarrow I = x \cdot \int {{{\sec }^2}xdx} - \int {\left[ {\dfrac{{dx}}{{dx}} \cdot \int {{{\sec }^2}xdx} } \right]dx}
⇒I=xtan⁡x−∫1⋅tan⁡xdx\Rightarrow I = x\tan x - \int {1 \cdot \tan xdx}
⇒I=xtan⁡x−∫tan⁡xdx\Rightarrow I = x\tan x - \int {\tan xdx}
⇒I=xtan⁡x−log⁡∣sec⁡x∣+C\Rightarrow I = x\tan x - \log \left| {\sec x} \right| + C
Thus, ∫xsec⁡2xdx=xtan⁡x−log⁡∣sec⁡x∣+C\int {x{{\sec }^2}xdx} = x\tan x - \log \left| {\sec x} \right| + C .

Note: The formula for integration by parts is given by ∫f(x)g(x)dx=f(x)∫g(x)dx−∫[ddxf(x)⋅∫g(x)dx]dx\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right) \cdot \int {g\left( x \right)dx} } \right]dx} .
In the formula for integration by parts, f(x)f\left( x \right) and g(x)g\left( x \right) are chosen by the method of ILATE.
ILATE is known as Inverse trigonometric functions, Logarithm functions, Algebraic functions, Trigonometric functions and Exponential functions.
The function f(x)f\left( x \right) must be a function lying before the function g(x)g\left( x \right) in the ILATE rule.
In the given question, we chose f(x)=xf\left( x \right) = x and g(x)=sec⁡2xg\left( x \right) = {\sec ^2}x , because x is an algebraic function and sec⁡2x{\sec ^2}x is an trigonometric function and according to ILATE rule algebraic functions lie before trigonometric functions.