Question

How to integrate \[2x{{\sec }^{2}}2xdx\]?

Answer 1

Hint: Write the given expression as \[\int{2x{{\sec }^{2}}2xdx}\] and take the constant term out of the integral. Now, to calculate the integral of \[x.{{\sec }^{2}}2x\] assume x as function 1 \[\left( {{f}_{1}}\left( x \right) \right)\] and \[{{\sec }^{2}}2x\] as function 2 \[\left( {{f}_{2}}\left( x \right) \right)\] and apply the rule of integration by parts given as: - \[\int{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)dx}=\left[ {{f}_{1}}\left( x \right).\int{{{f}_{2}}\left( x \right)dx} \right]-\int{\left\{ \left( \int{{{f}_{2}}\left( x \right)dx} \right).f_{1}^{'}\left( x \right) \right\}dx}\], to get the answer. Here, \[f_{1}^{'}\left( x \right)=\dfrac{d\left[ {{f}_{1}}\left( x \right) \right]}{dx}\]. Use the formula: - \[\int{{{\sec }^{2}}\left( ax+b \right)dx}=\dfrac{\tan \left( ax+b \right)}{a}\].

Complete step by step answer:
Here, we have been provided with the function \[2x{{\sec }^{2}}2x\] and we are asked to integrate it. Let us assume the integral as I, so we get,
\[\begin{align}
  & \Rightarrow I=\int{2x{{\sec }^{2}}2xdx} \\
 & \Rightarrow I=2\int{x{{\sec }^{2}}2xdx} \\
\end{align}\]
As we can see that we have to integrate \[x{{\sec }^{2}}2x\] which is a product of two functions. That means we have to apply integration by parts. Clearly, we can see that in the product \[x{{\sec }^{2}}2x\], ‘x’ is algebraic function and \[{{\sec }^{2}}2x\] is a trigonometric function. So, according to ILATE rule we have to assume x as function 1 \[\left( {{f}_{1}}\left( x \right) \right)\] and \[{{\sec }^{2}}2x\] as function 2 \[\left( {{f}_{2}}\left( x \right) \right)\]. Here, ILATE stands for: -
I \[\to \] Inverse trigonometric function
L \[\to \] Logarithmic function
A \[\to \] Algebraic function
T \[\to \] Trigonometric function
E \[\to \] Exponential function
So, the numbering of functions is done according to the order of appearance in the above list.
Now, to calculate the integral applying the formula: - \[\int{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)dx}=\left[ {{f}_{1}}\left( x \right).\int{{{f}_{2}}\left( x \right)dx} \right]-\int{\left\{ \left( \int{{{f}_{2}}\left( x \right)dx} \right).f_{1}^{'}\left( x \right) \right\}dx}\], we get,
\[\Rightarrow I=2\left[ \left( x\int{{{\sec }^{2}}2xdx} \right)-\left\{ \left( \int{{{\sec }^{2}}2xdx} \right).\dfrac{dx}{dx} \right\}dx \right]\]
Using the integral formula: - \[\int{{{\sec }^{2}}\left( ax+b \right)}=\dfrac{\tan \left( ax+b \right)}{a}\], we get,
\[\begin{align}
  & \Rightarrow I=2\left[ x.\dfrac{\tan 2x}{2}-\int{\dfrac{\tan 2x}{2}dx} \right] \\
 & \Rightarrow I=x\tan 2x-\int{\tan 2xdx} \\
\end{align}\]
Using the formula: - \[\int{\tan \left( ax+b \right)dx}=\dfrac{\ln \left| \sec \left( ax+b \right) \right|}{a}\], we get,
\[\Rightarrow I=x\tan 2x-\dfrac{\ln \left| \sec 2x \right|}{2}+C\], where ‘C’ is the constant of integration.

Note:
One may note that whenever we have to calculate the integral of the product of two different functions we have to apply the integration by parts method and numbering should be done according to the ILATE rule. So, you must remember the ILATE rule and the formula used for integration by parts method to solve the above question. You must remember the basic formulas of integration of all the trigonometric functions.